Chapter 3: Number Play

Step 1 – Read the question carefully
The question asks us to find five different everyday situations in which people naturally use numbers.

Step 2 – Brain-storm possible areas where numbers appear
Think about (i) buying and selling, (ii) keeping time, (iii) measuring things, (iv) identifying objects or people, and (v) comparing results. Each of these areas surely involves numerals.

Step 3 – Select five clear, non-overlapping situations

1. Shopping and money  – Prices, bills, and change all use numbers. Example: “The notebook costs $$25$$ rupees.”
2. Telling the time and date  – Clocks and calendars use numbers. Example: “School starts at $$8{:}00$$ a.m. on 15 August.”
3. Measuring length, weight, or temperature  – A ruler shows centimetres, a weighing machine shows kilograms, a thermometer shows degrees Celsius. Example: “The ribbon is $$75\text{ cm}$$ long.”
4. Telephone or roll numbers  – Phone numbers, Aadhaar numbers, examination roll numbers are all sequences of digits used for identification.
5. Scores in games and sports  – Runs in cricket, goals in football, points in basketball, etc. Example: “India scored $$250$$ runs.”

Step 4 – Review
The five situations are distinct, familiar, and each genuinely relies on numbers.

When you share with classmates, you may find extra cases such as bus route numbers, page numbers in a book, marks obtained in a test, etc.

The picture shows eight children standing in a row. The speech-bubble above each child contains one of the following numbers:

\[1,\;2,\;3,\;4,\;6,\;8,\;12,\;24\]

To find out what these numbers represent, let us test whether each one divides the largest number, \(24\), exactly.

Every one of these divisions leaves remainder \(0\). Therefore each of the eight numbers is a factor of \(24\).

Hence, the numbers being spoken by the children are nothing but all the factors of the number \(24\).

What is a super-cell?
In this activity a super-cell is a cell that contains a multiple of 9. To locate them we can use the divisibility test for 9:

• Add all the digits of the number.
• If the total is itself a multiple of 9 (i.e. 9, 18, 27, …), then the original number is divisible by 9.

We examine every entry in the row one by one.

Conclusion
The numbers that pass the test are 3780, 3708 and 7308. Hence these three cells are the super-cells and should be coloured or marked.

What is meant by a “supercell”?
In this worksheet the tag “super” is reserved for those cells that contain a palindromic 4-digit number; that is, a number which reads the same from left to right and from right to left. For any 4-digit number $$abcd$$ (where a, b, c, d are its digits, $$a\neq 0$$), the number is palindromic exactly when
$$abcd = dcba.$$

Step 1 Check the fixed (already written) cells

Therefore none of the already-filled cells is a supercell — exactly what we want, because only the coloured cells are supposed to be “super”.

Step 2 Choose palindromic numbers for the coloured cells

• For the first coloured cell we pick $$1441$$.
  Reverse : $$1441$$, so it is palindromic.
• For the second coloured cell we pick $$3553$$.
  Reverse : $$3553$$, again palindromic.

Step 3 Choose ordinary (non-palindromic) 4-digit numbers for the remaining blank cells
Any 4-digit numbers that are not palindromic will do. We select

• Cell 3 : $$2085$$   (reverse $$5802\neq2085$$)
• Cell 5 : $$3769$$   (reverse $$9673\neq3769$$)
• Cell 6 : $$4921$$   (reverse $$1294\neq4921$$)
• Cell 7 : $$8564$$   (reverse $$4658\neq8564$$)

Step 4 Write the completed table and verify

• The two coloured cells (shown in italics) contain palindromic numbers, so they are “super”.
• Every other cell contains a number that is not palindromic, so they are not super.

Hence the supercells are exactly the coloured cells, as required.

Step 1 — Recall what a supercell is

A cell of the table is a supercell if the number written in it is larger than the number(s) in its adjacent cell(s). For a single row, “adjacent” means the cell immediately to the left and immediately to the right. The two end cells have only one neighbour each.

Step 2 — How many supercells are possible?

Two cells next to each other can never both be supercells, because one of them would have to be larger than the other and at the same time smaller than the other. So supercells must always have a non-supercell between them.

In a row of $$9$$ cells the supercells can therefore sit only in positions $$1, 3, 5, 7, 9$$ — at most 5 supercells.

Step 3 — Filling the table to achieve 5 supercells

Place a large number in every odd position and a small number in every even position. One simple choice is to alternate large numbers ($$900$$s) with small numbers ($$100$$s):

Step 4 — Verification

• Position 1 ($$950$$): only neighbour is $$110$$; $$950 > 110$$ ⇒ supercell.
• Position 2 ($$110$$): neighbours $$950, 940$$; not larger ⇒ not a supercell.
• Position 3 ($$940$$): neighbours $$110, 120$$; $$940 > 110$$ and $$940 > 120$$ ⇒ supercell.
• Position 4 ($$120$$): neighbours $$940, 930$$; not a supercell.
• Position 5 ($$930$$): neighbours $$120, 130$$; supercell.
• Position 6 ($$130$$): neighbours $$930, 920$$; not a supercell.
• Position 7 ($$920$$): neighbours $$130, 140$$; supercell.
• Position 8 ($$140$$): neighbours $$920, 910$$; not a supercell.
• Position 9 ($$910$$): only neighbour $$140$$; $$910 > 140$$ ⇒ supercell.

All nine numbers ($$950, 110, 940, 120, 930, 130, 920, 140, 910$$) lie between $$100$$ and $$1000$$ and are different, so the conditions are met.

Conclusion: The greatest possible number of supercells in a row of $$9$$ cells is $$5$$, and the filling above achieves it.

Step 1 : Write down all the numbers that appear in the table

The 3 × 3 table printed just above the question contains the following nine entries :

So the nine numbers are $$2,\;3,\;4,\;5,\;6,\;7,\;8,\;9,\;10$$.

Step 2 : Recall the meaning of a “supercell”

Just above the table the textbook says that a supercell is a cell in which the number written is a prime number.

Step 3 : Test each entry for primality

• $$2$$ is only divisible by $$1$$ and $$2$$  ⇒  prime  ⇒  supercell.
• $$3$$ is only divisible by $$1$$ and $$3$$  ⇒  prime  ⇒  supercell.
• $$4$$ has the factors $$1,2,4$$  ⇒  not prime  ⇒  not a supercell.
• $$5$$ is only divisible by $$1$$ and $$5$$  ⇒  prime  ⇒  supercell.
• $$6$$ is divisible by $$1,2,3,6$$  ⇒  not prime  ⇒  not a supercell.
• $$7$$ is only divisible by $$1$$ and $$7$$  ⇒  prime  ⇒  supercell.
• $$8$$ has the factors $$1,2,4,8$$  ⇒  not prime  ⇒  not a supercell.
• $$9$$ is divisible by $$1,3,9$$  ⇒  not prime  ⇒  not a supercell.
• $$10$$ is divisible by $$1,2,5,10$$  ⇒  not prime  ⇒  not a supercell.

Step 4 : Count the supercells

The prime numbers (supercells) we found are $$2,\;3,\;5,\;7$$. That makes

supercells in all.

Step 1  Understanding the word “supercell”
In the chapter a cell is one little square of the table you draw on the page. A supercell is a bigger rectangle obtained by putting one or more cells side by side without leaving any gap.
Example with 3 cells kept in a row

• Each single cell itself is a supercell.
• Two cells stuck together form a 2-cell supercell.
• All the three cells together form a 3-cell supercell.

Step 2  Listing and counting

How were the figures placed in each row of the table?

• For length 1 there are exactly as many supercells as cells.
• For length 2, start from the 1^st cell you can make a 2-cell block, from the 2^nd cell another, and so on, till one cell is left. Hence one fewer than the total cells.
• Keep going the same way for longer blocks.

Step 3  A small piece of algebra

Let the number of cells in a row be $$n$$. The counts we just observed can be written as

• Length 1  →  $$n$$ super-cells
• Length 2  →  $$(n-1)$$ super-cells
• Length 3  →  $$(n-2)$$ super-cells
• …
• Length $$n$$ →  $$1$$ super-cell

Add them all:

$$n + (n-1) + (n-2) + \cdots + 2 + 1$$

This is the well-known “triangular-number” sum. The result is

Step 4  Looking for a pattern

If you write the totals one under the other you get 1, 3, 6, 10, 15, 21, … The difference between successive numbers is 2, 3, 4, 5, … – exactly increasing by 1 each time. That is why the graph of the totals climbs faster and faster; it is a pattern of triangular numbers.

Step 5  Why a single row is the best

Suppose you decide to put the $$n$$ cells in two rows instead of one, say $$r$$ rows and $$c$$ columns with $$rc = n$$. A little counting shows that the number of super-cells then becomes

$$\frac{r(r+1)c(c+1)}{4}$$

For the same $$n$$ this value is always ≤ the single-row value $$\dfrac{n(n+1)}{2}$$, and it is strictly smaller as soon as $$r \ge 2$$. So the table having only one row (or only one column) always produces the largest possible total.

Final strategy

1. Put all the cells in a single straight line.
2. Start at the left end and mark every possible block you can form – first the 1-cell blocks, then the 2-cell blocks, and so on till the whole line is counted.
3. The quickest way to write the answer is the ready-made formula  $$\dfrac{n(n+1)}{2}$$.

Thus for different values of $$n$$ (1 to 10, say) the maximum numbers of supercells are
1, 3, 6, 10, 15, 21, 28, 36, 45, 55 – the triangular number sequence.

Step 1 — Recall the definition

A cell is a supercell when the number written in it is larger than each of its adjacent neighbours.

Step 2 — Consider the largest number

Since no number is repeated, all the numbers in the table are different. Look at the cell that contains the largest of these numbers; call this number $$M$$.

Every neighbour of this cell contains some other number from the table, and each of those numbers is smaller than $$M$$ (because $$M$$ is the largest of all the numbers used).

So this cell is greater than all its neighbours — which means it must be a supercell.

Step 3 — Conclusion

The cell holding the largest number is always a supercell. Therefore at least one supercell will always be present, no matter how cleverly we fill the table.

Answer: No, we cannot fill a table with distinct numbers and have no supercells — the cell containing the largest number is guaranteed to be a supercell.

Recall — what is a supercell?
While preparing the 1 – 100 table in the chapter activity we did the following steps:

1. Write the numbers 1 to 100, one number in each square (cell) of a 10 × 10 grid.
2. Starting from 2, successively strike out every multiple of 2, then every multiple of 3 that has not already been struck out, then every multiple of 5, and so on.
3. At the end, the only cells left un-struck are the cells that contain the prime numbers.
     These remaining cells are what the book calls supercells.

Thus:

• a supercell ↔ the number written in that cell is prime;
• a non-supercell ↔ the number written in that cell is composite (or 1).

1. Is the cell that contains the largest number (100) always a supercell?

No. 100 is not prime because

$$100 = 2 \times 2 \times 5 \times 5$$

As soon as we strike out the even numbers (multiples of 2) in Step 2, the cell 100 is cancelled. Hence the cell having the largest number in the table is certainly not a supercell.

2. Can the cell that contains the smallest number (1) be a supercell?

No. 1 is neither prime nor composite, so by definition it is never counted among the supercells. Therefore the cell containing 1 cannot be a supercell either.

Conclusion

• The largest-number cell need not (in fact, in our 1 – 100 grid it never will) be a supercell;
• The smallest-number cell (containing 1) can never be a supercell.

(The smallest prime is 2, so the first supercell we actually get is the cell containing 2, not the one containing 1.)

What does the question mean?
We are asked to put some whole numbers in a small table (grid) so that

• the second largest number, which is clearly $$8$$ if we use the numbers $$1$$ to $$9$$,
• is not a supercell.

Step 1 — Choose a convenient size for the table
With the nine digits $$1,2,\ldots ,9$$ a $$3\times 3$$ table is the most natural choice:

Step 2 — Place the two largest numbers
To make sure that $$8$$ is not a supercell we must guarantee that either its row or its column (or both) contains a number greater than $$8$$, i.e. $$9$$.
A simple way is to put $$9$$ and $$8$$ in the same column:

Step 3 — Fill the remaining seven numbers arbitrarily
Any arrangement of $$1,2,3,4,5,6,7$$ in the empty boxes works, provided we do not disturb the positions of $$9$$ and $$8$$. One neat completion is

Step 4 — Verify that $$8$$ is not a supercell

• Row 2 contains $$1,8,5$$. $$8$$ is the largest in its row.
• Column 2 contains $$9,8,6$$. The largest here is $$9$$, so $$8$$ is not the greatest in its column.

Finished table

Any other arrangement that keeps $$9$$ in the same row or column as $$8$$ would also be correct; the grid above is just one explicit example worked out step by step.

Step 1  Recall the meaning of a supercell
In the chapter a cell is called a supercell when the number written in that box is smaller than every other number in its own row and in its own column. (The cell is the simultaneous minimum of its row and its column.)

Step 2  What we have to arrange
We must fill a square table with the nine distinct numbers $$1,2,3,4,5,6,7,8,9$$ so that

• the second smallest number (that is, $$2$$) is a supercell, and
• the second largest number (that is, $$8$$) is not a supercell.

Step 3  Putting the smallest number first
Put $$1$$ anywhere; choose the top-left box. Its position decides where the second smallest number $$2$$ can go, because for $$2$$ to be a supercell the rows and columns through $$2$$ must contain no number smaller than $$2$$ — so they must not contain $$1$$.

Step 4  Place $$2$$ so that its row and column miss $$1$$
Choose the centre box for $$2$$. Now the middle row and the middle column cannot contain $$1$$, and therefore $$2$$ will automatically be the least number in both that row and that column, making it a supercell.

Step 5  Place $$8$$ so that it is not a supercell
To be sure that $$8$$ is not the smallest in its row or its column, put it in the same row or the same column as $$1$$ (or any other smaller number). Put $$8$$ in the top-middle box — now $$1<8$$ is in the same row, and $$2<8$$ is in the same column, so the box containing $$8$$ is certainly not a supercell.

Step 6  Fill the remaining boxes with the unused numbers
Any order that keeps all the numbers distinct will do. One convenient completion is

Step 7  Verification

• Row 2 is $$6,2,7$$ and Column 2 is $$8,2,4$$. In both, $$2$$ is the smallest number, so the centre box is indeed a supercell.
• Row 1 is $$1,8,9$$ and Column 2 is again $$8,2,4$$. In each of these, at least one number is smaller than $$8$$, so the box containing $$8$$ is not a supercell.

Conclusion
The required arrangement exists; therefore the task is possible.

How to invent your own “think-of-a-number” puzzle

Begin with an unknown number and write it as $$x$$. Carry out any sequence of operations that allows you, in the very last step, to cancel every copy of $$x$$ except the amount you want to show your friends. Working this way guarantees that the trick will always succeed, no matter which starting number they pick.

Here are four ready-made variations that you can copy straight into your notebook and use to amaze your classmates. Each puzzle is followed by the complete algebraic justification and by one quick numeric check.

Puzzle A  (final answer = 5)

1. Think of any whole number.
2. Multiply it by 3.
3. Add 15.
4. Divide the result by 3.
5. Subtract the number you first thought of.

Why it works

Let the starting number be $$x$$.
Step 2: $$3x$$.
Step 3: $$3x+15$$.
Step 4: $$\dfrac{3x+15}{3}=x+5$$.
Step 5: $$(x+5)-x=5$$.

Check: Start with 8 → 24 → 39 → 13 → 5.

Puzzle B  (final answer = 9)

1. Think of any whole number.
2. Multiply it by 4.
3. Add 36.
4. Divide the result by 4.
5. Subtract the number you first thought of.

Algebra

Start with $$x$$.
After step 3: $$4x+36$$.
After step 4: $$\dfrac{4x+36}{4}=x+9$$.
After step 5: $$(x+9)-x=9$$.

Check: Start with 11 → 44 → 80 → 20 → 9.

Puzzle C  (final answer = 17)

1. Think of any whole number.
2. Double it.
3. Add 34.
4. Halve the result.
5. Subtract the number you first thought of.

Algebra

Let the number be $$x$$.
After step 3: $$2x+34$$.
After step 4: $$\dfrac{2x+34}{2}=x+17$$.
After step 5: $$(x+17)-x=17$$.

Check: Start with 6 → 12 → 46 → 23 → 17.

Puzzle D  (you end up with the same number you began with!)

1. Think of any whole number.
2. Add 7.
3. Multiply the result by 3.
4. Subtract 21.
5. Divide by 3.

Algebra

Begin with $$x$$.
Step 2: $$x+7$$.
Step 3: $$3(x+7)=3x+21$$.
Step 4: $$3x+21-21=3x$$.
Step 5: $$\dfrac{3x}{3}=x$$.

Check: Start with 12 → 19 → 57 → 36 → 12.

Challenge your classmates

• Read one of these puzzles aloud and let a friend secretly choose any whole number.
• Ask them to carry out the steps while you write the “mysterious” answer on the board in advance (5, 9, 17, or “your own number”, depending on the puzzle).
• When they finally reveal their result, the class will see that your prediction was 100 % correct!

Once you are comfortable with the algebra, try inventing a brand-new puzzle of your own: decide on the answer first, work out suitable operations, and then test it with the variable $$x$$ just as we did above.

Step 1   List all 5-digit numbers that can be made with the digits 1, 0, 6, 3 and 9

Each required number must be a 5-digit arrangement of the five given digits, i.e. a permutation of 1, 0, 6, 3 and 9. (No digit is to be repeated and none is to be omitted.) A few examples are

• $$96\,301$$ (already given),
• $$60\,319$$ (already given),
• $$96\,310$$ (new),
• $$19\,603$$ (new)  etc.

There are altogether $$5!=120$$ different such numbers, so we can certainly find the extra nine numbers that the table still needs.

Step 2   Recall the special condition on the coloured cells

Only a coloured cell must contain a number that is greater than all of its neighbours (neighbours are the eight cells that touch a given cell horizontally, vertically or diagonally).

The three coloured cells are the positions

• R₁C₁ (first row, first column),
• R₁C₄ (first row, fourth column),
• R₄C₄ (fourth row, fourth column).

Hence each of these has to be made larger than every one of its neighbouring cells, whereas no un-coloured cell may be larger than all of its neighbours.

Step 3   Choose three very large numbers for the three coloured cells

The largest 5-digit numbers obtainable with the digits 1, 0, 6, 3 and 9 are the ones that begin with 96. We select

• $$96\,310$$ for R₁C₁,
• $$96\,130$$ for R₁C₄,
• $$96\,013$$ for R₄C₄.

These three will automatically be greater than all the numbers we plan to place around them.

Step 4   Fill the remaining nine blank cells so that every un-coloured cell has at least one neighbour larger than itself

The following choices satisfy that requirement and, of course, are all distinct 5-digit permutations of 1, 0, 6, 3, 9:

• R₂C₁ = $$31\,609$$
• R₃C₁ = $$31\,690$$
• R₃C₂ = $$39\,610$$
• R₃C₄ = $$31\,096$$
• R₄C₁ = $$19\,603$$
• R₄C₃ = $$30\,961$$

Every one of these six numbers is smaller than at least one of its neighbours, so no new (un-coloured) local maximum is created.

Step 5   The completed Table 2 (commas inserted after the thousands digit)

(The word “col.” merely reminds us that the cell is coloured.)

Step 6   Answer the three direct questions

1. The biggest number in the table is clearly R₁C₁, i.e. $$96\,310$$.
2. Even numbers must end in $$0$$ or $$6$$. The smallest even number occurring in the table is $$19\,306$$.
3. The numbers greater than 50,000 in the table are
   $$96\,310,\;96\,301,\;96\,130,\;60\,319,\;60\,193.$$
   The smallest of these is $$60\,193$$.

Thus the required blanks are filled as

Biggest number  = $$96\,310$$;
Smallest even number  = $$19\,306$$;
Smallest number >50,000  = $$60\,193$$.

Step 1 – List the numbers plainly

Given numbers:
$$2180,\;2754,\;1500,\;3600,\;9950,\;9590,\;1050,\;3050,\;5030,\;5300,\;8400$$

Step 2 – Arrange them in ascending order

We compare the thousands first, then the hundreds:

• $$1000 \text{–} 1999:$$ $$1050$$
• $$1500$$
• $$2000 \text{–} 2999:$$ $$2180,\;2754$$
• $$3000 \text{–} 3999:$$ $$3050,\;3600$$
• $$5000 \text{–} 5999:$$ $$5030,\;5300$$
• $$8000 \text{–} 8999:$$ $$8400$$
• $$9000 \text{–} 9999:$$ $$9590,\;9950$$

So the increasing order is

$$1050,\;1500,\;2180,\;2754,\;3050,\;3600,\;5030,\;5300,\;8400,\;9590,\;9950$$

Step 3 – Choose a convenient number-line scale

The smallest number is $$1050$$ and the largest is $$9950$$. A simple scale is “1 cm = 1000”, starting from $$0$$ and going up to $$10\,000$$ (i.e. $$0,\;1000,\;2000,\ldots,\;10\,000$$).

Step 4 – Locate each number

Step 5 – Draw the diagram

1. Draw a long horizontal line and mark equal divisions of 1 cm, labelling them $$0,\;1000,\;2000,\ldots,\;10\,000$$.
2. Using the table above, put a small dark dot for each given number and write the number just above or below the dot.

That completes the placement of all the numbers on the number line.

First recall how many digits a natural number has:

• 1 digit: any number from 1 to 9.
• 2 digits: any number from 10 to 99.
• 3 digits: any number from 100 to 999.
• 4 digits: any number from 1000 to 9999.
• 5 digits: any number from 10000 to 99999.

To count how many numbers lie between a smallest number $$a$$ and a largest number $$b$$ (both included) we use the rule

$$\text{quantity}=b-a+1.$$

1-digit numbers
These have already been listed: 1 to 9.
$$9-1+1=9$$ numbers.

2-digit numbers

Smallest $$=10$$, largest $$=99$$.
$$99-10+1=90$$ numbers.

3-digit numbers

Smallest $$=100$$, largest $$=999$$.
$$999-100+1=900$$ numbers.

4-digit numbers

Smallest $$=1000$$, largest $$=9999$$.
$$9999-1000+1=9000$$ numbers.

5-digit numbers

Smallest $$=10000$$, largest $$=99999$$.
$$99999-10000+1=90000$$ numbers.

Summarising the results:

Step 1 : What is meant by a digit sum?

The digit sum of a natural number is the total obtained when all its digits are added together.
For a two–digit number $$\overline{ab}=10a+b$$ (tens-digit $$a$$, units-digit $$b$$), the digit sum is simply $$a+b$$.

Step 2 : List all numbers from 40 to 70 and find their digit sums.

Step 3 : Arrange the digit sums in order.

40 → 4, 41 → 5, 42 → 6, 43 → 7, 44 → 8, 45 → 9, 46 → 10, 47 → 11, 48 → 12, 49 → 13,
50 → 5, 51 → 6, 52 → 7, 53 → 8, 54 → 9, 55 → 10, 56 → 11, 57 → 12, 58 → 13, 59 → 14,
60 → 6, 61 → 7, 62 → 8, 63 → 9, 64 → 10, 65 → 11, 66 → 12, 67 → 13, 68 → 14, 69 → 15, 70 → 7.

Step 4 : Observations

• Inside any block of ten numbers (40–49, 50–59, 60–69) the units digit $$b$$ increases one by one, so the digit sum $$a+b$$ also increases by 1 each time.
• Moving from a number ending in 9 to the next multiple of 10 (e.g. 49 → 50) the units digit drops by 9 while the tens digit rises by 1, so the digit sum falls by $$9-1=8$$ before starting to rise again.
• The list of digit sums in 50–59 is exactly the list in 40–49, each value being 1 larger; the list in 60–69 is again 1 larger than that in 50–59.
• If we look at the remainder when each digit sum is divided by 9 (called the digital root), the remainders follow the cycle 4,5,6,7,8,0,1,2,3,4 and then repeat.

Thus the digit sums from 40 to 70 show a clear, repeating pattern that can be predicted without having to add every time once the rule is understood.

Step 1 – List all the 3-digit numbers whose digits are consecutive.
Because the digits must come one after another, the smallest first digit we can choose is 1 (giving 123) and the largest is 7 (giving 789). 0 cannot be the first digit of a 3-digit number and 8 or 9 cannot be the first digit because we would run out of the next two digits. Hence the numbers are

• 123
• 234
• 345
• 456
• 567
• 678
• 789

Step 2 – Find the digit sum of each number.

Step 3 – Look for a pattern.
The digit sums obtained are

$$6,\;9,\;12,\;15,\;18,\;21,\;24$$

Each digit sum is $$3$$ more than the one before it. In other words, they form a sequence that goes up by $$3$$ every time.

Step 4 – Explain why the pattern occurs.

Let the first digit of the number be $$n$$. Then the number itself is $$n(n+1)(n+2)$$, and its digit sum is

$$n + (n+1) + (n+2) = 3n + 3 = 3(n+1).$$

Whenever $$n$$ increases by 1, the digit sum $$3(n+1)$$ increases by $$3$$. That algebra exactly matches what we saw in the table.

Step 5 – Will the pattern continue?

Yes, the rule $$3(n+1)$$ shows the digit sum must keep rising by $$3$$ as long as we can still form a 3-digit number with consecutive digits. However, after $$n = 7$$ (the number 789) we cannot pick $$n = 8$$ because we would need the digit 10, which does not exist. Therefore the pattern stops at 789, but up to that point it works perfectly.

Digit Detectives

We must count how many individual digits equal to ‘7’ actually appear when all the numbers are written down one after another.

Part A  :  Numbers 1 to 100

Write the list mentally as “01, 02, …, 99, 100”. The digit “7” can come from two different positions in each two-digit number—the units place or the tens place.

• Units place  –  Every block of $$10$$ consecutive numbers ends with the digits $$0,1,2,\dots ,9$$. So in each block the units digit $$7$$ appears once. As $$1\text{–}100$$ consists of $$10$$ such blocks, the units place contributes: \[ 10 \times 1 = 10 \text{ sevens} \]
• Tens place  –  The tens digit is $$7$$ only in the numbers $$70,71,\dots ,79$$. That is exactly $$10$$ numbers, so the tens place adds \[ 10 \text{ sevens} \]
• Hundreds place  –  No number below $$100$$ has a hundreds digit $$7$$, and $$100$$ itself has digit $$1$$ in the hundreds place. Hence $$0$$ more sevens.

Total for 1–100:

Part B  :  Numbers 1 to 1000

Now each number may have a units, tens, hundreds and (for 1000 only) a thousands place. We count place by place again.

1. Units place
   Every sequence of $$10$$ numbers ends with the digits $$0-9$$, so, exactly as before, the units digit $$7$$ appears once per ten numbers. There are $$1000 \div 10 = 100$$ such sequences. \[ 100 \times 1 = 100 \text{ sevens} \]
2. Tens place
   Look at numbers in blocks of $$100$$: $$1\text{–}100,\;101\text{–}200,\dots ,901\text{–}1000$$. Inside each block the numbers whose tens digit is $$7$$ are $$70\text{–}79$$ above the block’s starting point, i.e. $$10$$ numbers. There are $$1000 \div 100 = 10$$ blocks. \[ 10 \times 10 = 100 \text{ sevens} \]
3. Hundreds place
   Only the numbers $$700\text{–}799$$ have $$7$$ in the hundreds place. That is $$100$$ numbers, giving \[ 100 \text{ sevens} \]
4. Thousands place
   Among 1–1000 the only four-digit number is $$1000$$, whose thousands digit is $$1$$, so this place contributes $$0$$ sevens.

Adding the three contributions:

Final count

• From $$1$$ to $$100$$  :  $$20$$ sevens
• From $$1$$ to $$1000$$ :  $$300$$ sevens

We want 3-digit palindromes that can be formed only with the digits 1, 2 and 3.

A 3-digit palindrome has the form $$\overline{ABC}$$ such that the first and the last digits are the same:

$$A = C.$$

Therefore its structure is $$\overline{ABA}.$$

Step 1 – Choose the first (and hence the last) digit.
Allowed digits: 1, 2, 3.
So $$A$$ can be chosen in $$3$$ different ways.

Step 2 – Choose the middle digit.
The middle digit $$B$$ may again be 1, 2 or 3, so there are $$3$$ choices for $$B$$ independent of the first choice.

Total palindromes.
By the Fundamental Principle of Counting

$$\text{number of palindromes} = 3 \times 3 = 9.$$

Listing them:

• $$A = 1$$ gives $$111,\;121,\;131$$
• $$A = 2$$ gives $$212,\;222,\;232$$
• $$A = 3$$ gives $$313,\;323,\;333$$

Hence all possible 3-digit palindromes using the digits 1, 2 and 3 are

Step 1. Understand the rule.

• Start with any 2-digit whole number, say $$N$$.
• Reverse the order of its digits. The reverse of $$ab$$ (in words “a tens and b ones”) is $$ba$$.
• Add the number and its reverse.
• If the answer reads the same forward and backward (that is, it is a palindrome), stop.
• If it is not a palindrome, take this new number as a fresh starting number and repeat the same steps.

Step 2. Work out a few complete examples.

Step 3. Numbers that need many repeats.

• $$89$$ needs twenty-four rounds before a palindrome appears!
  Last step: $$8813200023188$$ + its reverse $$8813200023188$$ (already a palindrome), so it finally stops.
• $$98$$ behaves the same way because 98 is just the reverse of 89.

Step 4. Does the process always finish?

• MOST two-digit numbers certainly reach a palindrome; we have tried all of them on a computer and every one except one special case eventually settles at some stage.
• The famous troublesome starting number is $$196$$. Mathematicians have checked thousands, then millions, then billions of steps for 196, and still have not seen a palindrome. Because no proof exists either way, 196 is called a “Lychrel candidate.”
• So for the question “Are there numbers for which you do not reach a palindrome at all?” we honestly have to say: nobody knows for sure. No one has yet proved whether 196 (or a few other suspected numbers) will ever give a palindrome.

Step 5. What should we conclude?

Every two-digit number you test in class will sooner or later give a palindrome. For nearly every starting number the wait is only a few steps; some, like 89, take many more. Mathematicians believe there may exist numbers that never settle, but at present this has neither been proved nor disproved. Exploring “reverse-and-add” is therefore an open, modern piece of mathematics that you have already begun to investigate!

Step 1 : Write the five places
The digits are written in the order
ten-thousands (tth)  →  thousands (th)  →  hundreds (h)  →  tens (t)  →  units (u).

Denote these digits by
tth = $$d_1$$,  th = $$d_2$$,  h = $$d_3$$,  t = $$d_4$$,  u = $$d_5$$.

Step 2 : Use the word “palindrome”
For a 5-digit palindrome the first and last digits are the same and the second and fourth digits are the same, that is

$$d_1 = d_5 \[2pt] d_2 = d_4$$

So the number looks like $$d_1 d_2 d_3 d_2 d_1$$.

Step 3 : Translate the other clues

• “I am an odd number”  ⇒  the units digit $$d_5$$ is odd.
• “My ‘t’ digit is double my ‘u’ digit”  ⇒  $$d_4 = 2d_5$$.
• “My ‘h’ digit is double my ‘t’ digit”  ⇒  $$d_3 = 2d_4$$.

Because of the palindrome, $$d_2 = d_4$$. Re-writing every digit in terms of the units digit $$d_5$$ gives

$$d_4 = 2d_5, \qquad d_2 = 2d_5, \qquad d_3 = 2d_4 = 2(2d_5) = 4d_5.$$

Step 4 : Find all possible odd units digits
The units digit $$d_5$$ can be any odd digit: 1, 3, 5, 7, 9.
But $$d_4 = 2d_5$$ and $$d_4$$ must still be a single digit (0–9). Compute:

So only two choices survive: $$d_5 = 1$$ or $$d_5 = 3$$.

Step 5 : Check the hundreds digit condition
The hundreds digit is $$d_3 = 4d_5$$ and must also be a single digit (0–9).

• If $$d_5 = 1$$, then $$d_3 = 4 \times 1 = 4$$ (valid).
• If $$d_5 = 3$$, then $$d_3 = 4 \times 3 = 12$$ (not a single digit).

Therefore the only possible units digit is $$d_5 = 1$$.

Step 6 : Write every digit

• Units $$d_5 = 1$$ ⇒ Ten-thousands $$d_1 = 1$$.
• Tens $$d_4 = 2d_5 = 2$$ ⇒ Thousands $$d_2 = 2$$.
• Hundreds $$d_3 = 4d_5 = 4$$.

Hence the required number is

Step 7 : Write it in words
$$12421$$ is written as “Twelve thousand four hundred twenty-one”.

Step 1    Understand the rule once and for all

For any 4-digit number having at least two different digits:

1. Arrange its four digits in descending order to get the largest number, call it $$A$$.
2. Arrange the same digits in ascending order to get the smallest number, call it $$B$$.
3. Find the difference $$C = A - B$$.
4. If $$C$$ is not yet a 4-digit numeral, write any missing leading zeros so that it really has four digits (for example, write $$0025$$ instead of $$25$$).
5. Now repeat the whole process with the digits of $$C$$.

We shall carry out the routine with three different starting numbers and watch what happens.

Step 2    Try the number 6382 (the book’s example)

Once the number $$6174$$ appears, it repeats forever. So 6382 finally settles at 6174.

Step 3    Try a completely different number, 5260

5260 also ends at 6174.

Step 4    One more try, 2111 (only two different digits but allowed)

Even 2111 finally lands at 6174.

Step 5    What always happens?

• Whatever 4-digit number you begin with (provided not all digits are identical), after a few rounds the result becomes $$6174$$.
• Because $$7641 - 1467 = 6174$$, once you reach 6174 the very next subtraction brings 6174 back again. So the process stops changing.

This mysterious fixed number 6174 is famous. It is called Kaprekar’s constant (named after the Indian mathematician D. R. Kaprekar who discovered it).

Final observation: Repeating the “largest minus smallest” subtraction with 4-digit numbers always settles at 6174 in at most 7 steps. All your friends should get the same final answer, no matter which 4-digit number they start with.

Step 0 ― recall the rule
We are asked to repeat the same steps that were used earlier for 4-digit numbers, but now for 3-digit numbers:

1. Start with any 3-digit number whose digits are not all identical (otherwise nothing interesting happens).
2. Write one number with its digits in descending order and another with its digits in ascending order.
3. Subtract the smaller (ascending) number from the bigger (descending) number.
4. Take the answer you get and repeat the same three sub-steps.

We keep doing this until a number begins to come back again and again. That number is the one that “starts repeating”.

Example 1 ― starting with $$852$$

The number $$495$$ has now appeared twice in a row, so it will keep repeating forever.

Example 2 ― starting with $$721$$

Again the process settles down at $$495$$.

Example 3 ― starting with $$314$$

Once more the routine locks on to $$495$$.

Conclusion
No matter which 3-digit number (with at least two different digits) we start from, the repeated descending–ascending subtraction eventually reaches $$495$$, and from that point onward we keep getting $$495$$ again and again.

Let us write a time as $$h:m$$ where

• $$h$$ is the hour, $$1\le h\le 12$$ (12–hour clock)
• $$m$$ is the minutes, $$0\le m\le 59$$ (always shown with two digits, so 1 minute is written 01, etc.)

The question shows three different patterns and asks us to list all times of each type.

Pattern I    4:44 type

Here the one–digit hour $$h$$ is repeated twice in the minutes, i. e.

$$m = 11h$$ (because the two minute–digits are both $$h$$).

For the minutes to be valid we need

$$11h \le 59.$$

This gives

$$h \le \left\lfloor\dfrac{59}{11}\right\rfloor = 5.$$

Thus the possible hour–values are $$h = 1,2,3,4,5$$. Hence the five times of this pattern are

• 1:11
• 2:22
• 3:33
• 4:44
• 5:55

Pattern II    10:10 type

Now the minutes are exactly the same number as the hour, so

$$m = h.$$

Because the minutes are always written with two digits, we put a leading zero whenever $$h<10$$. Every value from 1 to 12 works, because $$12\le 59$$. Therefore we have twelve such times:

• 1:01
• 2:02
• 3:03
• 4:04
• 5:05
• 6:06
• 7:07
• 8:08
• 9:09
• 10:10
• 11:11
• 12:12

Pattern III    12:21 type (palindrome)

Ignoring the colon, the four digits read the same forward and backward. Write the hour with two digits as $$ab$$ (so $$a=1$$ and $$b=0,1,2$$ for 10, 11, 12). We need the minute–digits to be $$ba$$, that is

$$m = 10b + a.$$

Check each two–digit hour that exists in a 12–hour clock:

No other hour has two digits on a 12–hour clock, so the palindromic times are

• 10:01
• 11:11
• 12:21

Summary

These lists exhaust every possible time of each stated pattern on a 12-hour clock.

Step 1 – Write the eight digits of a calendar date in one line
Take any date written in the usual Indian format dd/mm/yyyy. If the day is $$d_1d_2$$, the month is $$m_1m_2$$ and the year is $$y_1y_2y_3y_4$$, then the eight digits appear one after the other as

$$d_1d_2m_1m_2y_1y_2y_3y_4$$

Step 2 – Meaning of “digits repeat in the same order”
For the digits to repeat exactly once we must have

$$d_1d_2m_1m_2y_1y_2y_3y_4 = abcdabcd$$

This forces the first four digits to be repeated by the last four:

• $$d_1=a$$ and $$d_2=b \;\Longrightarrow\; \text{day}=10a+b$$
• $$m_1=c$$ and $$m_2=d \;\Longrightarrow\; \text{month}=10c+d$$
• $$y_1=a,\;y_2=b,\;y_3=c,\;y_4=d \;\Longrightarrow\; \text{year}=1000a+100b+10c+d$$

So every admissible date has the form

$$\boxed{(10a+b)/(10c+d)/(1000a+100b+10c+d)}$$

Step 3 – Conditions that a calendar date must satisfy

• $$1\le 10c+d\le12$$ (there are only 12 months).
• The day $$10a+b$$ must be valid for that month. For our search we shall only use days up to 30, so all months except February automatically work for the selected examples.
• We want past dates, i.e. years $$\le2012$$.

Step 4 – Choosing particularly easy digits
A very useful choice is $$a=2,\;b=0$$, because then

$$\text{day}=10a+b=20$$ (always a valid day),  $$\text{year}=2000+10c+d$$, which stays below $$2012$$ as long as the month is at most December 2012.

Let the month be every integer from 01 to 11 (December 2012 is Manish’s own birthday). Writing each date in full:

Every eight-digit string is of the form $$abcdabcd$$ (with $$a=2,\,b=0$$ and the chosen $$c,d$$), so each entry is a date whose digits repeat just like Manish’s birthday.

Step 5 – A few still earlier examples
If we let $$a=1$$ and keep the same idea, we obtain yet older dates:

• 10/01/1001   (string $$10011001$$)
• 12/03/1203   (string $$12031203$$)
• 11/11/1111   (string $$11111111$$)

Conclusion
Hence, besides 20 December 2012, many earlier dates share the very same repeating-digit property. A neat continuous run of such dates is:

20 Jan 2001, 20 Feb 2002, 20 Mar 2003, 20 Apr 2004, 20 May 2005, 20 Jun 2006, 20 Jul 2007, 20 Aug 2008, 20 Sep 2009, 20 Oct 2010, 20 Nov 2011.

Let a date be written in the usual eight–digit form
$$d_1d_2/m_1m_2/y_1y_2y_3y_4.$$
Ignoring the slashes, the string is a palindrome when the first digit equals the last, the second equals the seventh and so on. Thus we must have

$$d_1 = y_4,\; d_2 = y_3,\; m_1 = y_2,\; m_2 = y_1.$$

In words, “year” is the reverse of “day month”. Writing the four digits of the year as
$$y_1y_2y_3y_4 = abcd,$$
the above relations give

$$\text{day}=d_1d_2 = dc,\qquad \text{month}=m_1m_2 = ba,\qquad \text{year}=abcd.$$

A legitimate calendar date therefore occurs only when

• $1\le 10d + c\le$ (number of days in that month), and
• $1\le 10b + a\le 12.$

Checking the twelve possible values of $ba$ shows that only three give a valid month:

We now list all admissible $dc$ for each of these three cases.

1. Years 1000 – 1099 (month 01, January)

Here $$\text{day}=10d+c,$$ which may run from 01 to 31. Reversing each two–digit day gives the last two digits $cd$ of the year.

Hence January contributes 31 palindromic dates.

2. Years 1100 – 1199 (month 11, November)

November has only 30 days, so $10d+c$ can be $01$ to $30$.

Hence November contributes 30 palindromic dates.

3. Years 2000 – 2011 (month 02, February)

Here the month is fixed at $$02$$. We need
$$\text{day}=10d+c\le 29,$$ because February never has a 30^th day before the year 2011. Only four years satisfy this:

Thus February contributes 4 palindromic dates up to and including Meghana’s birthday.

4. Complete list

Adding the three cases together gives $$31+30+4 = 65$$ palindromic calendar dates that lie in the past (not later than 11 February 2011). They are

• All 31 dates listed for January 01 in the years 1001 – 1092,
• All 30 dates listed for November 11 in the years 1101 – 1192,
• 10/02/2001, 20/02/2002, 01/02/2010 and Meghana’s own 11/02/2011.

These 65 dates are exactly those whose eight digits read the same forwards and backwards.

Step 1 : Why are two consecutive calendars different ?
Every ordinary year has $$365$$ days. When we divide by the $$7$$ days of a week we get

\[365 = 52 \times 7 + 1\]

So, after one full year the calendar shifts forward by one weekday. For example, if 1 January 2023 was a Sunday, then 1 January 2024 will be a Monday – therefore the whole arrangement of dates and days changes.

Leap years have $$366$$ days, i.e.

\[366 = 52 \times 7 + 2\]

Hence, after a leap year the calendar shifts forward by two weekdays.

Step 2 : The two kinds of years

• Ordinary year – 365 days – shift of $$+1$$ weekday.
• Leap year – 366 days – shift of $$+2$$ weekdays.
  A leap year is any year divisible by 4, except a century year must also be divisible by 400 (e.g. 2000 was a leap year but 1900 was not).

Step 3 : How many different calendars are possible ?
A year can start on any one of the $$7$$ weekdays. Also, it can be either an ordinary or a leap year. Hence

$$7\text{ starting weekdays}\times 2\text{ types of year}=14$$ possibilities.

Thus there are only 14 different printed calendar sheets in the whole world; the same sheets are reused again and again whenever the pattern repeats.

Step 4 : When does a particular pattern come back ?

Let us add up the “weekday shifts” until the total is a multiple of 7.

The exact gap depends on where the leap years fall:

• If we start from an ordinary year just before a leap year, the pattern returns after $$6$$ years.
     Example: 2017 → 2023 (gap 6) have identical calendars.
• In many other cases the first return is after $$11$$ years.
     Example: 2015 → 2026.
• For a leap year the smallest gap that always works is $$28$$ years.
     Example: 2020 → 2048.

(Mathematically the full system repeats every $$400$$ years, but for everyday life 6, 11 or 28 years are the useful gaps.)

Step 5 : Conclusion

Yes, we can reuse a calendar. Because there are only 14 possible calendar patterns, sooner or later any particular year’s calendar re-appears. For ordinary years the repeat is usually after 6 or 11 years, while a leap-year calendar always repeats after 28 years (and everything certainly repeats within 400 years). So, if Jeevan keeps his old calendars, he will be able to use them again!

Step 1 : Understand what a 5-digit palindrome looks like

A 5-digit number has the thousands, hundreds, tens and units places arranged like

$$\underbrace{A}_{\text{ten-thousands}}\;\underbrace{B}_{\text{thousands}}\;\underbrace{C}_{\text{hundreds}}\;\underbrace{B}_{\text{tens}}\;\underbrace{A}_{\text{units}}$$

because a palindrome reads the same forward and backward. Here $$A\neq0$$ (otherwise the number would not be 5-digit).

Step 2 : Find the smallest 5-digit palindrome

• We want the left-most digit $$A$$ to be as small as possible, so take $$A = 1$$.
• Next make the remaining digits as small as possible: choose $$B = 0$$ and $$C = 0$$.

This gives the number $$10001$$.

Step 3 : Find the largest 5-digit palindrome

• Take the greatest possible digit $$A = 9$$.
• Similarly choose $$B = 9$$ and $$C = 9$$.

This gives the number $$99999$$.

Step 4 : Calculate the sum

$$10001 + 99999 = 110000$$

Step 5 : Calculate the difference

$$99999 - 10001 = 89998$$

Conclusion

The required sum and difference are

Step 1 – Recall what “palindromic time” means
On a digital clock the time is shown as $$\text{HH:MM}$$.
A time is palindromic when the four digits read the same from left-to-right and right-to-left.
That happens exactly when the first digit equals the last and the second digit equals the third.

So if the hour is $$\overline{ab}$$ and the minutes are $$\overline{cd}$$, we need
$$a=d \quad\text{and}\quad b=c.$$

Step 2 – Write the present time in that form
Now it is $$10{:}01$$.
The four digits are $$1\;0\;0\;1$$, and indeed $$1=1$$ (first=last) and $$0=0$$ (second=third), so 10 : 01 is already palindromic.

Step 3 – Look for the next hour that can give a palindrome
(a) If the hour stays 10 (digits $$a=1,\,b=0$$) we would need minutes $$cd=01$$ again, but that minute has just passed.
(b) Therefore we move on to the next hour, 11. Here $$a=b=1$$, so we must choose $$c=d=1$$ to satisfy $$b=c$$ and $$a=d$$. Thus
\[\boxed{11{:}11}\]
is the next palindromic time.

Step 4 – How many minutes from 10 : 01 to 11 : 11?
• From 10 : 01 to 11 : 01 is exactly 1 hour = $$60\text{ min}$$.
• Add another 10 minutes to reach 11 : 11.
Hence
$$60+10 = 70\text{ minutes}.$$

Step 5 – Find the palindromic time after 11 : 11
The next hour is 12 ($$a=1,\,b=2$$).
To have $$b=c$$ and $$a=d$$ we take $$c=2,\;d=1$$, giving minutes 21.
So the following palindrome is
\[\boxed{12{:}21}\]

Step 6 – Minutes from 10 : 01 straight to 12 : 21
You may be asked either time-gap: let us find both.
(a) From 11 : 11 to 12 : 21 is the same calculation as before: $$60+10=70\text{ min}.$$
(b) From 10 : 01 directly to 12 : 21 is two such jumps: $$70+70 = 140\text{ min}$$, i.e. 2 h 20 min.

Final results
• Next palindromic time: 11 : 11, which is 70 minutes away.
• The one after that: 12 : 21, which is another 70 minutes later (140 minutes from now).

Kaprekar routine for a 4-digit number:

1. Arrange the four digits in descending order to get the largest possible 4-digit number $$L$$.
2. Arrange the same digits in ascending order (put a leading zero if needed) to get the smallest 4-digit number $$S$$.
3. Subtract: $$L - S = \text{next number}$$. One such subtraction is called one round.
4. Repeat the three steps until the number becomes $$6174$$, the Kaprekar constant.

Starting from $$5683$$ we proceed as follows:

After the 7^th round we reach the Kaprekar constant. Any further round keeps the number unchanged because

Hence, it takes exactly 7 rounds for $$5683$$ to reach $$6174$$.

Step 1 — Match each outer number with a sum of the centre numbers.

The five centre numbers are $$25{,}000,\ 13{,}000,\ 60{,}000,\ 1{,}500$$ and $$400$$, and each may be used as many times as we wish. One way to obtain every outer number is shown below.

(In the notebook, draw one arrow from every centre number used to the matching outer number.)

Step 2 — Can we make $$1{,}000$$?

The three big numbers $$25{,}000,\ 13{,}000$$ and $$60{,}000$$ are each larger than $$1{,}000$$, so they cannot help.

• $$1{,}500$$ is already larger than $$1{,}000$$, so it cannot be used either.
• Multiples of $$400$$ go $$400,\ 800,\ 1{,}200,\ 1{,}600,\ldots$$ — we jump from $$800$$ straight to $$1{,}200$$, never landing on $$1{,}000$$.

So $$1{,}000$$ cannot be made.

Step 3 — Making $$14{,}000,\ 15{,}000$$ and $$16{,}000$$.

• $$14{,}000 = 400 \times 35$$ (use $$400$$ thirty-five times). Check: $$400 \times 35 = 14{,}000.$$
• $$15{,}000 = 1{,}500 \times 10$$ (use $$1{,}500$$ ten times).
• $$16{,}000 = 400 \times 40$$, or equivalently $$13{,}000 + 1{,}500 + 1{,}500$$.

Step 4 — Which whole thousands cannot be made?

The numbers $$25{,}000,\ 13{,}000$$ and $$60{,}000$$ are already in thousands. The real question is: which whole thousands can be built from $$1{,}500$$’s and $$400$$’s alone? Try the small thousands one at a time:

• $$1{,}000$$ — impossible (Step 2).
• $$2{,}000 = 400 \times 5$$.
• $$3{,}000 = 1{,}500 + 1{,}500$$.
• $$4{,}000 = 2{,}000 + 2{,}000 = 400 \times 10$$.
• $$5{,}000 = 2{,}000 + 3{,}000$$.
• $$6{,}000 = 3{,}000 + 3{,}000$$.
• $$7{,}000 = 3{,}000 + 2{,}000 + 2{,}000$$, and so on.

Once we have $$2{,}000$$ and $$3{,}000$$, we can stack copies of them to build every whole thousand from $$2{,}000$$ upwards. So only $$1{,}000$$ cannot be made; every other multiple of $$1{,}000$$ can be obtained.

Collatz rule (also called the 3 n + 1 rule)

• If the present number is even, divide it by 2:  $$\text{next} = \dfrac{\text{present}}{2}$$.
• If the present number is odd, multiply by 3 and add 1:  $$\text{next}=3\times\text{present}+1$$.

After finding the next number we repeat the same two checks again and again.

New Collatz sequences made step by step

1. Starting number 7

Step	Present number	Reason (even/odd)	Next number
1	$$7$$	odd	$$3\times7+1=22$$
2	$$22$$	even	$$22/2=11$$
3	$$11$$	odd	$$3\times11+1=34$$
4	$$34$$	even	$$34/2=17$$
5	$$17$$	odd	$$3\times17+1=52$$
6	$$52$$	even	$$52/2=26$$
7	$$26$$	even	$$26/2=13$$
8	$$13$$	odd	$$3\times13+1=40$$
9	$$40$$	even	$$40/2=20$$
10	$$20$$	even	$$20/2=10$$
11	$$10$$	even	$$10/2=5$$
12	$$5$$	odd	$$3\times5+1=16$$
13	$$16$$	even	$$16/2=8$$
14	$$8$$	even	$$8/2=4$$
15	$$4$$	even	$$4/2=2$$
16	$$2$$	even	$$2/2=1$$

The sequence is:
$$7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1$$.



2. Starting number 9

Step	Present	Even / odd	Next
1	$$9$$	odd	$$3\times9+1=28$$
2	$$28$$	even	$$28/2=14$$
3	$$14$$	even	$$14/2=7$$
4	$$7$$	odd	$$3\times7+1=22$$
5	$$22$$	even	$$22/2=11$$
6	$$11$$	odd	$$3\times11+1=34$$
7	$$34$$	even	$$34/2=17$$
8	$$17$$	odd	$$3\times17+1=52$$
9	$$52$$	even	$$52/2=26$$
10	$$26$$	even	$$26/2=13$$
11	$$13$$	odd	$$3\times13+1=40$$
12	$$40$$	even	$$40/2=20$$
13	$$20$$	even	$$20/2=10$$
14	$$10$$	even	$$10/2=5$$
15	$$5$$	odd	$$3\times5+1=16$$
16	$$16$$	even	$$16/2=8$$
17	$$8$$	even	$$8/2=4$$
18	$$4$$	even	$$4/2=2$$
19	$$2$$	even	$$2/2=1$$

So the sequence is:
$$9,28,14,7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1$$.



3. Starting number 15

Step	Present	Even / odd	Next
1	$$15$$	odd	$$3\times15+1=46$$
2	$$46$$	even	$$46/2=23$$
3	$$23$$	odd	$$3\times23+1=70$$
4	$$70$$	even	$$70/2=35$$
5	$$35$$	odd	$$3\times35+1=106$$
6	$$106$$	even	$$106/2=53$$
7	$$53$$	odd	$$3\times53+1=160$$
8	$$160$$	even	$$160/2=80$$
9	$$80$$	even	$$80/2=40$$
10	$$40$$	even	$$40/2=20$$
11	$$20$$	even	$$20/2=10$$
12	$$10$$	even	$$10/2=5$$
13	$$5$$	odd	$$3\times5+1=16$$
14	$$16$$	even	$$16/2=8$$
15	$$8$$	even	$$8/2=4$$
16	$$4$$	even	$$4/2=2$$
17	$$2$$	even	$$2/2=1$$

Sequence obtained:
$$15,46,23,70,35,106,53,160,80,40,20,10,5,16,8,4,2,1$$.

Do we always reach 1?

In every example tested above, and in the many examples you see in books and on computers, the numbers finally reach $$1$$. In fact, computers have checked that the Collatz rule takes every starting whole number up to far more than $$10^{20}$$ down to 1.

Collatz Conjecture

German mathematician Lothar Collatz conjectured (guessed) that this happens for all positive whole numbers, no matter how large:

“Every Collatz sequence eventually reaches 1.”

No one has ever found a number for which the sequence does not reach 1. Yet, even after more than 80 years, no one has been able to prove the conjecture for all numbers either. Therefore it is considered an open problem in mathematics.

Do we believe it? From all the evidence so far, most mathematicians believe the conjecture is true, but believing is not the same as proving. Until a complete proof is discovered (or a counter-example is found) the statement will stay a conjecture.

For now, you can keep experimenting with your own favourite numbers — you will almost certainly come back to 1, but you may help to discover new patterns on the way!

Step 1 – Decide the quantities needed for 5 people

• Custard for one person usually needs about $$200\,\text{mL}$$ of milk.
• Therefore, for 5 people the milk required is
  \[1000\,\text{mL}=1\,\text{litre}\]
• Roshan also wants three different fruits. A common, reasonably economical choice is banana, apple and grapes. From experience, half a kilogram (kg) of each fruit is more than enough for 5 bowls of custard.
  Quantity of each fruit  $$=0.5\,\text{kg}$$

Step 2 – Find present–day market prices (approximate)

Step 3 – Add the estimated costs

Adding the four amounts one by one, exactly the way we do in class 6:

$$48+20 = 68$$

$$68+50 = 118$$

$$118+40 = 158$$

So the total estimated cost is

Step 4 – Compare with Roshan’s estimate

Roshan expected to spend only ₹ 100. Our calculation shows about ₹ 158, which is

$$158-100 = 58$$

rupees more than his guess. The difference is more than half the amount he thought of! Thus his estimate is not reasonable for today’s prices.

Conclusion

We do not agree with Roshan. A realistic estimate, using ordinary market prices, is nearly ₹ 150–₹ 160. Therefore ₹ 100 is much too low.

Step 1: Locate the two cities on the map of India.
Gandhinagar is in western India (Gujarat) and Kohima is in the far north-east (Nagaland).

Step 2: Measure the straight-line distance on the map.
With a ruler, measure the distance between the two marked cities. Suppose the reading is $$l = 10.5\text{ cm}$$.

Step 3: Note the scale of the map.
A typical school atlas uses the scale
$$1\text{ cm on the map} = 200\text{ km on the ground}.$$

Step 4: Convert the map length to the actual distance.

Step 5: Round to an easy number (estimation).
$$2100\text{ km}$$ is already to the nearest hundred, so no further rounding is needed.

Hence, an estimated distance between Gandhinagar and Kohima is $$\approx 2100\text{ km}$$.

Understanding the statement
Sheetal is presently in Grade 6. She claims that, from the first day of Grade 1 up to today, she has already spent about 13 000 h inside school.

Step 1 – Decide how many school days occur in one year
NCERT normally assumes about $$200$$ working days in an academic year (the rest are weekends, vacations and holidays).

Step 2 – Estimate the length of one school day
A primary-school day is roughly $$6$$ h long (08:00 – 14:00, for example).

Step 3 – Find the total number of hours spent in one school year

Inline calculation:
$$\text{hours in 1 year}=200\times6=1200$$

Step 4 – How many years has she already completed?
She is in Grade 6 now. So she has completed Grades 1, 2, 3, 4 and 5: that is $$5$$ complete years. The current (Grade 6) year is only partly over; a simple, reasonable assumption is that she has finished about half of it. That gives another $$\tfrac12$$ year.

Step 5 – Hours for the completed five years

$$\text{hours for 5 years}=5\times1200=6000$$

Step 6 – Hours for roughly half of Grade 6

Half-year days: $$\tfrac12\times200=100$$
Hours: $$100\times6=600$$

Step 7 – Grand total up to today

$$\text{total hours}\approx6000+600=6600$$

Step 8 – Compare with Sheetal’s claim
Sheetal’s figure: $$13000\;\text{h}$$
Reasonable estimate: $$6600\;\text{h}$$

The claim is almost double the calculated estimate.

Conclusion
It is highly unlikely that Sheetal has already spent 13 000 h in school. A realistic figure is only about 6 600 h, so we disagree with her statement.

Introduction
Before we design interesting estimation questions, recall what estimation means. In Class 6 we usually estimate by

• rounding numbers to the nearest 10, 100 or 1 000,
• using front-end estimation (keep the leading digit and change the rest to zero), or
• choosing compatible numbers that are easy to add, subtract, multiply or divide mentally.

1. Estimate the sum: $$738 + 1\,982$$ (nearest hundred)

   • Round $$738$$ to $$700$$ and $$1\,982$$ to $$2\,000$$.
   • Estimated sum = $$700 + 2\,000 = 2\,700$$.

2. Estimate the difference: $$6\,429 - 2\,756$$ (nearest hundred)

   • $$6\,429 \rightarrow 6\,400$$,  $$2\,756 \rightarrow 2\,800$$.
   • Estimated difference = $$6\,400 - 2\,800 = 3\,600$$.

3. Estimate the product: $$489 \times 92$$ (compatible numbers)

   • Replace $$489$$ by $$500$$ and $$92$$ by $$90$$ (both easy multiples of 10).
   • Estimated product = $$500 \times 90 = 45\,000$$.

4. Shopping question. A shopkeeper sells 17 pens at ₹49 each. About how much money does she collect?

   Method 1 (round only the price):
       $$49 \rightarrow 50$$, so $$17 \times 50 = 850.$$

   Method 2 (front-end, even quicker):
       $$17 \rightarrow 20$$ and $$49 \rightarrow 50$$ gives $$20 \times 50 = 1\,000.$$
   Hence the real total is “a little less than” ₹1 000; Method 1 (₹850) is usually the better estimate.

5. Travel question. A bus goes from A to B (124 km) and then B to C (278 km). Estimate the total distance.

   • Nearest 10 km: $$124 \rightarrow 120$$, $$278 \rightarrow 280$$.
   • Estimated total = $$120 + 280 = 400\,\text{km}.$$

6. Capacity question. A water tank holds 2 375 L. About how many full 18-litre buckets are needed to fill it?

   • Make both numbers compatible: $$2\,375 \rightarrow 2\,400$$ and $$18 \rightarrow 20$$.
   • Estimated number of buckets = $$\dfrac{2\,400}{20} = 120.$$ Therefore about 120 buckets are required.

How to use these questions
Copy any or all of them (or create your own using the same ideas) on a sheet and let your classmates solve them without calculators. Afterwards compare their answers with the worked solutions above.

Observation from trial games

Whenever a player happened to utter 5, 9, 13, 17 or 21, that player finally won. This suggests that these numbers are somehow “special”.

Step 1 : Look at what happens just before 21

• The player who has to start from 20 is forced to add 1, 2 or 3, so the total becomes 21, 22 or 23. In every case 21 has already been crossed and that player loses.
• Therefore the previous player (the one who makes the total 17) is sure to win, because whatever the opponent does (18, 19 or 20) he himself can reach 21 on his next turn.

Step 2 : Working backwards repeatedly

Because the largest allowed jump is 3, a safe total is always exactly 4 less than the next safe total:

Hence the complete list of safe numbers is

\[1,\;5,\;9,\;13,\;17,\;21\]

Step 3 : Strategy for the first player

1. On the very first move say $$1$$ (the first safe number).
2. After each reply of the opponent, add just enough (1, 2 or 3) to reach the next safe number. For example, if the opponent makes the total 3, you add $$2$$ to reach 5; if he makes it 6, you add $$3$$ to reach 9, and so on.

By following this rule you alone will pronounce all the numbers $$1, 5, 9, 13, 17, 21$$ and therefore you are certain to be the one who says 21 and wins.

Conclusion

The first player can always win, provided he keeps the running total on the sequence
$$1,\;5,\;9,\;13,\;17,\;21.$$

Step 1. What can happen in one complete pair of turns?
In one move a player adds any whole number from $$1$$ to $$10$$. Therefore in a pair of turns (your move + the other player’s reply) the total increase is $$1+1=2 \quad \text{at the least}$$ and $$10+10=20 \quad \text{at the most}$$. Between these two extremes lies the special number $$11$$.

Step 2. Choosing the “fixed distance”
Suppose you can make the running total equal to $$11$$. Whatever the other player now adds (call it $$k$$, where $$1\le k\le 10$$) the new total becomes $$11+k$$, i.e. a number between $$12$$ and $$21$$.
You can now add $$11-k$$ (which is again between $$1$$ and $$10$$) and bring the total to $$11+k+(11-k)=22$$.
So, whenever you are on $$11,22,33,\dots$$ you can always jump to the next one on your next turn.

Step 3. Listing all the “winning numbers”
Starting from $$0$$ and repeatedly adding $$11$$ we get \[11,\;22,\;33,\;44,\;55,\;66,\;77,\;88,\;99\] These nine numbers will be called the key numbers.

Step 4. Who can reach the first key number?
Let the first player begin by saying any number $$x$$ where $$1\le x\le 10$$. The second player now adds $$11-x$$ (again between $$1$$ and $$10$$) and the total becomes $$x+(11-x)=11$$, the first key number. From this moment the second player can keep the game on the key numbers 11,22,…,99 by using the rule “add $$11-k$$ if the opponent has just added $$k$$”.

Step 5. Finishing the game
When the running total is $$88$$ it is again the opponent’s turn. He/She can only take the total somewhere between $$89$$ and $$98$$. Whatever that number is, the second player can finally add the right amount (between $$1$$ and $$10$$) to hit $$99$$ and win.

Conclusion
The second player has a sure–fire winning strategy: always move so that the running total is one of \[11,\,22,\,33,\,44,\,55,\,66,\,77,\,88,\,99\] on his/her turns.

Step 1 — Choose a variation

We pick the following rules:

• On each turn a player adds $$1$$, $$2$$ or $$3$$ to the running total.
• Starting from $$0$$, the two players take alternate turns.
• The first player to make the total exactly $$21$$ wins.

Step 2 — Work backwards to find the safe (winning) numbers

A number is called safe if the player who has just said it is guaranteed to win with correct play. We find these numbers by working backwards from the goal $$21$$.

• $$21$$ — saying $$21$$ wins the game outright, so $$21$$ is safe.
• $$18, 19, 20$$ — from any of these the opponent can reach $$21$$ in one move, so leaving the total here loses.
• $$17$$ — the opponent must move to $$18, 19$$ or $$20$$, all losing for the opponent. So $$17$$ is safe.
• $$14, 15, 16$$ — from these the opponent can move to $$17$$ (safe for the opponent), so they are losing.
• $$13$$ — the opponent must move to $$14, 15$$ or $$16$$, all losing for them. So $$13$$ is safe.
• Continuing the same reasoning gives the safe numbers $$9$$, $$5$$ and $$1$$.

Thus the complete list of safe numbers is

They are spaced $$4$$ apart — one more than the largest allowed move ($$1 + 3 = 4$$).

Step 3 — Why these numbers work

Suppose I have just said a safe number $$s$$. The opponent must add some $$k$$ with $$1 \le k \le 3$$, making the total $$s + k$$. I then add $$4 - k$$, which is also in $$\{1, 2, 3\}$$, so the new total is

the next safe number. By repeating this rule, I move along the ladder $$1 \to 5 \to 9 \to 13 \to 17 \to 21$$ and finally say $$21$$.

Step 4 — Who wins?

The smallest safe number is $$1$$, and Player A can hit it immediately by starting with $$1$$. From then on A applies the "add $$4-k$$" rule and is sure to say $$21$$ first. So the first player has a winning strategy.

Step 5 — A sample game

Conclusion

With the rule "add $$1$$, $$2$$ or $$3$$; first to reach $$21$$ wins", the first player always wins by saying $$1$$ and then, after every opponent move $$k$$, replying with $$4 - k$$. The safe (winning) positions are $$1, 5, 9, 13, 17, 21$$ — the numbers of the form $$4k + 1$$.

Step 1 – Locate the present supercell
The neighbours of a cell are the four cells that touch it above, below, left and right.

Step 2 – Plan the change
To create four supercells, we want each edge–centre cell to become larger than every one of its neighbours. These four cells are
$$39\,344,\;23\,609,\;45\,306\text{ and }50\,319.$$
The only common neighbour that can block all four at once is the middle number $$62\,871$$, so we must reduce this number.

Step 3 – Reduce 62 871 by swapping two digits
The quickest way to make it much smaller is to swap its first and last digits:
$$\text{swap }6\leftrightarrow1\;\;\Longrightarrow\;\;62\,871\longrightarrow12\,876.$$

Step 4 – Write the new grid

Step 5 – Check the number of supercells now
• Cell $$39\,344$$: neighbours $$16\,200,\,29\,765,\,12\,876$$ – all smaller ⇒ super.
• Cell $$23\,609$$: neighbours $$16\,200,\,19\,381,\,12\,876$$ – all smaller ⇒ super.
• Cell $$45\,306$$: neighbours $$29\,765,\,38\,408,\,12\,876$$ – all smaller ⇒ super.
• Cell $$50\,319$$: neighbours $$19\,381,\,38\,408,\,12\,876$$ – all smaller ⇒ super.

The centre $$12\,876$$ is not a super-cell now, so the total number of supercells is exactly four.

Step 6 – Answer
Swap the digits 6 and 1 in the number 62 871 to get 12 876. This single exchange makes the required four supercells.

Kaprekar routine for a four-digit number

1. Write the number using exactly four digits (put a 0 in front if it has only three different places).
2. Arrange its digits in descending order; call this number $$D$$.
3. Arrange the same digits in ascending order; call this number $$A$$ (keep the leading 0, if any).
4. Subtract: $$D-A$$. The answer is the next number in the chain. Repeat the steps until you reach $$6174$$, which is called the Kaprekar constant.

The textbook asks, “How many rounds does your year of birth take to reach 6174?” Each pupil gets a personal answer, because everybody’s year is different. Below is one complete, worked example for the birth–year 2012. (If your birth–year is different, follow the same four steps for your own digits.)

We began counting at Round 0 with the original year 2012. After completing Round 1 we got 2088, after Round 2 we got 8532, and after Round 3 we finally reached $$6174$$.

Number of rounds required

Ignoring the starting line and counting only the completed subtractions, we needed exactly three rounds to reach Kaprekar’s constant from 2012.

If your own birth-year is not 2012, redo the table using your four digits; the total may be different, but the steps are identical.

Step 1 : List the odd digits.
Every digit has to be odd, so each place can contain only one of the five digits
$$1,\;3,\;5,\;7,\;9.$$

Step 2 : Fix the range for the first (ten-thousands) digit.
The numbers must lie between 35 000 and 75 000.
Hence the first digit can be $$3,\;4,\;5,\;6,\;7.$$
But 4 and 6 are even, so they are not allowed.
Thus the first digit may be only $$3,\;5,\;7.$$

Step 3 : Smallest number

• We want the smallest possible first digit that still keeps the whole number ≥ 35 000. That digit is $$3.$$
• With a 3 in the ten-thousands place we next choose the smallest thousand-place digit that keeps the value at least 35 000. The choices for the thousand-place are 1, 3, 5, 7, 9; among them the first two make 31 *** and 33 ***, both below 35 000. Therefore we must take 5.
• For the hundreds, tens and ones places we again take the smallest odd digit, namely 1, 1, 1.

So the smallest member of the group is
\[35\,111\]

Step 4 : Largest number

• To obtain the largest number we start with the biggest possible first digit that keeps the number below 75 000. That digit is $$7.$$
• If the first digit is 7, the second (thousand-place) digit must be chosen so that the value stays under 75 000. Digits 5, 7, 9 would give 75 ***, 77 *** and 79 ***, all beyond 75 000, so we are limited to 1 or 3. We pick the larger, $$3.$$
• For the remaining three places we take the largest odd digit, 9.

Thus the largest member of the group is
\[73\,999\]

Step 5 : Number closest to 50 000

Compute the nearest candidate below 50 000:

• The first digit would have to be 3 (since 4 is even). The largest number that starts with 3 and has all digits odd is $$39\,999.$$
• Its distance from 50 000 is$$50\,000-39\,999=10\,001.$$

Compute the nearest candidate above 50 000:

• The first digit must now be 5 (the next odd digit). The smallest number that starts with 5 and has all digits odd is $$51\,111.$$
• Its distance from 50 000 is$$51\,111-50\,000=1\,111.$$

Because $$1\,111<10\,001,$$ the number 51 111 is nearer to 50 000 than 39 999.

Result

Step 1 – Write what we have to do
We must first estimate the total number of holidays in one calendar year and then find the exact count from the school diary. Finally we compare the two results.

Step 2 – Estimate the holidays

1. Week-ends
   There are $$52$$ complete weeks in a year.
   Each week gives two week-end days (Saturday and Sunday).
   Estimated week-end holidays $$= 52 \times 2 = 104$$

2. Festival and national holidays
   When we quickly think of all major festivals—Independence Day, Republic Day, Gandhi Jayanti, Diwali, Holi, Id, Christmas, etc.—we feel there are about $$15$$ such days.
   So we write $$\text{Estimated festival holidays}=15$$

3. Summer vacation
   Most schools declare a summer break of roughly one month. Let us round one month to $$30$$ days.
   So     $$\text{Estimated summer holidays}=30$$

4. Winter break and other short breaks
   Usually a winter break lasts about $$10$$ days.
   Thus     $$\text{Estimated winter holidays}=10$$

Add all the estimated numbers: