Chapter 6: Number Play

Given children are again standing in a line (exactly as in the Grade 6 textbook). Each child now shows a plane figure and, instead of naming the figure, calls out only one number. The numbers called out are, in order,

$$0,\;0,\;1,\;2,\;2,\;1$$

We have to find out what these numbers tell us.

Recall from Grade 6
The same picture (with the same six figures) appeared in Chapter 13 of the Class 6 text while discussing symmetry. At that time the children called out the names of the figures. The six figures, from left to right, are

1. a scalene triangle,
2. a general (non323 rectangle) parallelogram,
3. an isosceles trapezium,
4. a rectangle,
5. a rhombus,
6. a kite.

Step 1 – Write the number of lines of symmetry of each figure

Step 2 – Compare with the numbers called out

Arranging the results in the same left-to-right order gives

$$0,\;0,\;1,\;2,\;2,\;1$$

This is precisely the list that the children shout in the figure.

Conclusion

The number each child calls out is the number of lines of symmetry (axes of mirror symmetry) possessed by the plane figure that child is holding.

Step 1 – What does each child represent?
In the activity each child is wearing a card that has one digit written on it. Thus every child actually represents a digit of a multi-digit number.

Originally the digits on the seven cards were

$$6,\;5,\;2,\;2,\;1,\;0,\;0$$

(You can verify from the picture in the book that these are the digits lying on the ground next to the children.)

Step 2 – Re-arranging the children
The teacher now asks them to stand in ascending order of the digits written on their cards (smallest digit first, largest digit last). After this rearrangement the order of the cards becomes

$$0,\;0,\;1,\;2,\;2,\;5,\;6$$

Exactly these digits are the numbers you hear the children call out in the figure.

Step 3 – What number do these digits form?
Placing the digits side by side, as the children are now standing, gives the string
$$0012256$$
However, a number cannot start with the digit $$0$$. To write the smallest proper 7-digit number that can be formed with all seven given digits, we keep the first non-zero digit $$1$$ in the ten-lakh place and put the two zeros just after it. Thus we obtain

Step 4 – Interpreting the spoken digits
The spoken digits mean “We have arranged the seven given digits in ascending order; together they form the smallest 7-digit number that can be written with those very digits, namely $$1002256$$.”

Therefore the numbers 0, 0, 1, 2, 2, 5, 6 are simply the digits of that smallest possible number after the children have lined themselves up in increasing order of their individual digits.

Given observation

The page shows the six numbers

$$142857,\;285714,\;428571,\;571428,\;714285,\;857142$$

You are asked: “Could you figure out what these numbers convey? Observe and try to find out.”

Step 1 – List the digits in each number

Each of the six numbers uses exactly the same six digits –

$$1,\;4,\;2,\;8,\;5,\;7$$

and every digit appears once in every number. The order, however, keeps changing (or cycling).

Step 2 – Check if the numbers are multiples of one common number

Take the first (smallest) number as a trial divisor:

$$142857$$

Multiply it successively by the natural numbers $$1,2,3,4,5,6$$ and write the products.

Exactly the six numbers on the page appear as those six products. Therefore each given number is a multiple of $$142857$$.

Step 3 – Notice the cyclic rearrangement

Look carefully at the products. Reading each product from left to right, the digits keep shifting to the left while the left-most digit goes to the right end. For example

• $$142857$$ → move 1 step → $$285714$$
• $$285714$$ → move 1 step → $$428571$$

This is called a cyclic permutation of the digits 1 4 2 8 5 7.

Step 4 – Connect with the fraction $$\tfrac17$$

Divide 1 by 7 to two or three repetitions:

\[\frac17 = 0.142857\,142857\,142857\ldots\quad(repeating)\]

The six-digit block that repeats is exactly $$142857$$ – the same number we just multiplied. When this block repeats once, twice, thrice, … we obtain its cyclic rearrangements, which are precisely the given six numbers.

Concluding observation

The six numbers are the six non-zero multiples of $$142857$$ that are less than $$10^6$$, and they show the repeating block in the decimal expansion of $$\tfrac17$$ cycling through its digits. Hence the numbers “convey” the remarkable pattern hidden in the fraction $$\tfrac17$$.

In other words:

• Each number = $$142857 \times n$$ for $$n = 1,2,3,4,5,6$$,
• They are cyclic rearrangements of one another,
• All of them come from the repeating decimal of $$\tfrac17$$.

That is the hidden message in the set of numbers on the page.

Given arrangement (from the child standing right in front of the teacher to the one at the far end of the line)

Each child has to call out a number according to the rule:

• Standing at position $$i$$, count all the children at positions $$1,2,\dots ,(i-1)$$ who are taller than the child at position $$i$$.
• The count itself is what the child at position $$i$$ says aloud.

Let the height of the child in the $$i^{\text{th}}$$ position be $$h_i$$ (in cm). The number to be called out by that child is therefore

$$N_i = \bigl|\{\,j \mid 1 \le j < i \text{ and } h_j > h_i\}\bigr|.$$

We now compute $$N_i$$ for every position.

1. Anita (position 1)
   No one is standing in front of her, so the set in the above definition is empty and
   $$N_1 = 0.$$
2. Babu (position 2)
   Check only Anita: $$h_1 = 143 < h_2 = 155,$$ so no taller child. Hence $$N_2 = 0.$$
3. Charu (position 3)
   Children in front: Anita (143 cm), Babu (155 cm). Only Babu is taller than Charu (149 cm). Therefore $$N_3 = 1.$$
4. Dev (position 4)
   Heights in front: 143 cm, 155 cm, 149 cm — all smaller than Dev’s 165 cm. Thus $$N_4 = 0.$$
5. Esther (position 5)
   Heights in front: 143 cm, 155 cm, 149 cm, 165 cm. Only 165 cm is taller than Esther's 158 cm  $$\Rightarrow 1$$ such child, so $$N_5 = 1.$$
6. Faisal (position 6)
   Heights in front: 143 cm, 155 cm, 149 cm, 165 cm, 158 cm. Taller children: 155 cm, 165 cm, 158 cm  $$\Rightarrow 3$$ of them. Hence $$N_6 = 3.$$
7. Gita (position 7)
   All six children in front are shorter than 172 cm, so $$N_7 = 0.$$

Putting the results together, the list of numbers the children call out (in the same order as they are standing) is:

The available cards show the numbers $$3, 5, 7, 9, 11, 13$$. Notice that every one of these numbers is odd.

1. What happens when we add odd numbers?

• odd $$+$$ odd $$=$$ even
• even $$+$$ odd $$=$$ odd

So, as we keep adding odd numbers, the result flips between even and odd each time we add another odd number.

2. Add five odd numbers step by step.

Start with the first two cards:

odd $$+$$ odd $$=$$ even

Add a third card:

even $$+$$ odd $$=$$ odd

Add a fourth card:

odd $$+$$ odd $$=$$ even

Add the fifth card:

even $$+$$ odd $$=$$ odd

Therefore, the sum of any five odd numbers is itself odd.

3. Compare with the required total.

The puzzle demands a total of $$30$$.

$$30 = 2 \times 15$$ is an even number.

4. Final conclusion.

The sum of five odd cards must be odd, but $$30$$ is even. Hence it is impossible to place five cards chosen from $$3, 5, 7, 9, 11, 13$$ in the boxes so that they add to $$30$$.

So the puzzle has no solution with the given cards.

Step 1 – Recall the definition of an even number
A whole number is called even if it can be written in the form $$2k$$, where $$k$$ is some other whole number.

Step 2 – Represent each even addend algebraically
Assume we are going to add $$n$$ even numbers. Because each is even, we may write them as

• first even number  = $$2k_1$$
• second even number = $$2k_2$$
• third even number = $$2k_3$$
• \(\vdots\)
• nth even number = $$2k_n$$

Here $$k_1,\,k_2,\,\ldots ,\,k_n$$ are all whole numbers (0, 1, 2, 3, …).

Step 3 – Add the even numbers

Their sum is

Step 4 – Recognise the nature of the result

Because $$k_1 + k_2 + k_3 + \dots + k_n$$ is itself a whole number (call it $$K$$), equation (1) can be rewritten as

But any number of the form $$2K$$ is even by definition. Therefore the sum of the $$n$$ even numbers is even.

Step 5 – Does the count of addends matter?
The working above never required a particular value of $$n$$ — it only used that each addend is even. Hence, whether you add 2, 3, 4, or any number $$n$$ of even numbers, the final result is always even.

Step 6 – Numerical check (optional)

• $$8 + 4 = 12$$ (even)
• $$6 + 10 + 14 = 30$$ (even)
• $$2 + 18 + 20 + 26 = 66$$ (even)

Every trial confirms the algebraic proof.

Step 1 – Recall what an odd number looks like algebraically
Every odd whole number can be written in the form $$2k+1$$, where $$k$$ is some whole number (0, 1, 2, 3, …).

Step 2 – Add two odd numbers
Let the two odd numbers be $$2k+1$$ and $$2m+1$$. Then
$$(2k+1)+(2m+1) = 2k+2m+2 = 2(k+m+1).$$
The right-hand side is 2 times a whole number, so the result is even.
Conclusion: The sum of two odd numbers is even.

Step 3 – Add three odd numbers
Add a third odd number $$2n+1$$:
$$(2k+1)+(2m+1)+(2n+1) = 2k+2m+2n+3 = 2(k+m+n+1)+1.$$
This is one more than an even number, so the total is odd.
Conclusion: The sum of three odd numbers is odd.

Step 4 – General rule for $$r$$ odd numbers
Suppose we add $$r$$ odd numbers $$(2k_1+1),\,(2k_2+1),\,\dots,\,(2k_r+1)$$. Their sum is
$$(2k_1+1)+(2k_2+1)+\dots+(2k_r+1) = 2(k_1+k_2+\dots+k_r)+r.$$
The first part $$2(k_1+k_2+\dots+k_r)$$ is always even, so the parity of the whole sum depends only on $$r$$:

• If $$r$$ is even, the term $$r$$ is even, so the whole sum is even + even = even.
• If $$r$$ is odd, the term $$r$$ is odd, so the whole sum is even + odd = odd.

Final conclusion

If an even number of odd numbers are added, the sum is even. If an odd number of odd numbers are added, the sum is odd. So, yes — it does matter how many odd numbers you add.

Let us write the three odd numbers in their general algebraic form.

An odd number is always of the shape $$2k+1$$ where $$k$$ is any whole number. Take three such numbers:

$$2m+1, \; 2n+1, \; 2p+1$$  (where $$m,n,p$$ are whole numbers)

Add them:

$$\begin{aligned} (2m+1)+(2n+1)+(2p+1) &=2m+2n+2p+3\\[2pt] &=2(m+n+p+1)+1. \end{aligned}$$

The expression $$2(m+n+p+1)+1$$ is again of the form $$2K+1$$ (with $$K=m+n+p+1$$), so the sum is odd.

What does this mean for pairing?

• To arrange objects "in pairs" each pair must contain exactly 2 objects, so the total number of objects must be an even number.
• An odd total always leaves one object unpaired.

Since the sum of three odd numbers is itself odd, it cannot be completely arranged in pairs – one item will always be left over.

Concrete check: $$1+3+5=9$$ (odd); try making pairs out of 9 sticks – the 9th stick is left out.

Therefore the answer is: No, the sum of three odd numbers cannot be arranged entirely in pairs.

Key idea: Every odd whole number can be expressed in the algebraic form $$2k+1$$, where $$k$$ is an integer (0, 1, 2, 3, …).

We examine the sum of four, five and six such numbers separately.

(a) Sum of 4 odd numbers

Let the four odd numbers be

$$2k_1+1, \; 2k_2+1, \; 2k_3+1, \; 2k_4+1$$

Add them:

$$\begin{aligned} (2k_1+1) &+ (2k_2+1) + (2k_3+1) + (2k_4+1) \\ &= 2(k_1+k_2+k_3+k_4) + 1+1+1+1 \\ &= 2(k_1+k_2+k_3+k_4) + 4 \\ &= 2\bigl(k_1+k_2+k_3+k_4+2\bigr) \;(\text{factor 2}) \end{aligned}$$

The final expression is 2 multiplied by an integer, so the sum is an even number.

(b) Sum of 5 odd numbers

Let the five odd numbers be

$$2m_1+1, \; 2m_2+1, \; 2m_3+1, \; 2m_4+1, \; 2m_5+1$$

Add them:

$$\begin{aligned} &\;(2m_1+1)+(2m_2+1)+(2m_3+1)+(2m_4+1)+(2m_5+1) \\ &= 2(m_1+m_2+m_3+m_4+m_5)+5 \\ &= 2(m_1+m_2+m_3+m_4+m_5)+4+1 \\ &= 2\bigl(m_1+m_2+m_3+m_4+m_5+2\bigr)+1 \end{aligned}$$

This is of the form $$2q+1$$, again with $$q$$ an integer. Therefore the result is an odd number.

(c) Sum of 6 odd numbers

Let the six odd numbers be

$$2n_1+1, \; 2n_2+1, \; 2n_3+1, \; 2n_4+1, \; 2n_5+1, \; 2n_6+1$$

Add them:

$$\begin{aligned} &\;(2n_1+1)+(2n_2+1)+(2n_3+1)+(2n_4+1)+(2n_5+1)+(2n_6+1) \\ &= 2(n_1+n_2+n_3+n_4+n_5+n_6)+6 \\ &= 2\bigl(n_1+n_2+n_3+n_4+n_5+n_6+3\bigr) \end{aligned}$$

The expression is divisible by 2, so the sum is an even number.

Conclusion

The pattern we have just proved matches a well-known general rule:

• The sum of an even number of odd numbers is always even.
• The sum of an odd number of odd numbers is always odd.

Specifically for this question we obtained:

\[\text{4 odd numbers} \;\Longrightarrow\; \textbf{even}\]
\[\text{5 odd numbers} \;\Longrightarrow\; \textbf{odd}\]
\[\text{6 odd numbers} \;\Longrightarrow\; \textbf{even}\]

Step 1 – Introduce variables
Let the younger sibling’s present age be $$x\text{ years}$$.
Since they were born exactly one year apart, the elder sibling’s age is $$x+1\text{ years}$$.

Step 2 – Translate the statement into an equation
Maria says the sum of their ages is 112: 
$$x + (x+1) = 112$$

Step 3 – Solve the equation
Combine like terms:
$$2x + 1 = 112$$
Subtract 1 from both sides:
$$2x = 111$$
Divide by 2:
$$x = 55.5$$

Step 4 – Interpret the result
$$x = 55.5$$ means the younger sibling would be 55 years 6 months old, which is impossible when ages are measured in complete years on a birthday.

Step 5 – Reason with parity (even–odd check)
An even quicker check is to notice that the ages differ by 1 year, so they are consecutive integers. For any integer $$n$$:

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

Since $$2n+1$$ is always odd, the sum of two consecutive ages must be odd. But 112 is even, so such ages cannot exist.

Conclusion
Therefore, Maria’s claim is not possible. Their ages cannot add up to 112.

Step 1 – Introduce variables
Let

• $$x$$ be the number of ₹1 coins,
• $$y$$ be the number of ₹5 coins,
• $$z$$ be the number of ₹10 coins.

According to the statement:

• $$x$$ is odd,
• $$y$$ is odd,
• $$z$$ is even.

Step 2 – Write the total-value equation
The money in the piggy bank is

$$1\times x + 5\times y + 10\times z = 205.$$

Step 3 – Express the parity (even–odd) of each term
Because

• an odd number of ₹1 coins contributes an odd amount,
• 5 × (odd) is also odd (for example, 5, 15, 25, …),
• 10 × (even) is even (every multiple of 10 formed with an even integer ends in 0 and is divisible by 2),

the sum of the three contributions is

odd + odd + even = even.

Step 4 – Compare with the given total
But 205 is an odd number, not an even one. Therefore the calculated total conflicts with the parity we just obtained, so it cannot be correct.

Algebraic check (optional but instructive)

Let odd numbers be written as $$2k+1$$ and even numbers as $$2k$$:

• $$x = 2m + 1,$$
• $$y = 2n + 1,$$
• $$z = 2p.$$

Then

$$\begin{aligned} \text{Total} &= x + 5y + 10z \\ &= (2m+1) + 5(2n+1) + 10(2p) \\ &= 2m + 1 + 10n + 5 + 20p \\ &= 2(m + 5n + 10p + 3).\end{aligned}$$

The total equals 2 × (some whole number), so it must be even. Hence it can never be 205.

Conclusion
Lakpa did make a mistake; under the given conditions no selection of coins can total ₹205.

Step 1 · Recall what “parity” means
“Parity” tells us whether a whole number is even or odd.
• An even number is divisible by 2, i.e. $$k=2q$$ for some whole $$q$$.
• An odd number is not divisible by 2, i.e. $$k=2q+1$$ for some whole $$q$$.

Step 2 · Write the number of small squares as a product
If a rectangular grid has $$m$$ rows and $$n$$ columns of small squares, then the total number of small squares is
\[\text{total}=m\times n\].

Step 3 · State the rule for the parity of a product
For any two whole numbers $$m$$ and $$n$$:

• If both $$m$$ and $$n$$ are odd, their product $$m\times n$$ is odd.
• If either $$m$$ or $$n$$ is even (or both are even), the product $$m\times n$$ is even.

This rule comes directly from the fact that multiplying by an even number always introduces a factor 2.

Step 4 · Decide the parity without multiplying
Therefore, to know whether the number of small squares is odd or even you need only look at each dimension:

Step 5 · Example check
Suppose the grid is 9 × 6.

• 9 is odd, 6 is even ⇒ at least one factor is even ⇒ the product is even.
• No actual multiplication is required to conclude this.

Conclusion
You can tell the parity of the total number of small squares just by checking whether each dimension is odd or even; only when both are odd will the total also be odd, otherwise it is even.

Goal. Construct an algebraic expression whose value is always even, no matter what whole number we substitute.

Key fact. A number is even  ↔  it can be written as $$2k$$ for some integer $$k$$.

Step 1 – Choose an even coefficient.
Let the variable be $$n$$.  14 is even, so the product

$$14n$$

is even because $$14n = 2(7n)$$.

Step 2 – Add (or subtract) another even number.
Adding or subtracting an even number keeps the result even, since

even ± even = even.

Pick $$-8$$ (which is even). The new expression is

Step 3 – Verify explicitly.

• Rewrite $$14n - 8$$ as $$2(7n - 4)$$.
• The bracket $$7n - 4$$ is an integer for every integer $$n$$.
• Hence the whole value has the form $$2k$$, proving it is always even.

Conclusion. The expression $$14n - 8$$ always has even parity.

Many other answers are possible, for example: $$6p$$, $$18q + 10$$, $$50r - 4$$, and so on.

Key idea : An integer is odd precisely when it can be written in the form $$2k+1$$ for some integer $$k$$. Therefore an algebraic expression that can always be rearranged to $$2(\text{an integer})+1$$ is guaranteed to have odd parity.

Below are four different expressions, each proved to be odd for every integer value of the variable.

1. Expression A: $$2n+1$$

   Write any integer $$n$$ in one of two ways.

   • If $$n$$ is even, say $$n=2k$$, then $$2n+1=2(2k)+1=4k+1=2(2k)+1$$ which is odd.
   • If $$n$$ is odd, say $$n=2k+1$$, then $$2n+1=2(2k+1)+1=4k+3=2(2k+1)+1$$ which is also odd.

   Thus $$2n+1$$ is always odd.

2. Expression B: $$4n+3$$

   Factor a 2:

   $$4n+3=2(2n+1)+1$$

   The bracket $$2n+1$$ is an integer for every integer $$n$$, so the entire right-hand side is of the form $$2k+1$$ and hence odd.

3. Expression C: $$2n^2+2n+1$$

   Factor the first two terms:

   $$2n^2+2n+1=2n(n+1)+1$$

   The product $$n(n+1)$$ involves two consecutive integers, one of which is always even, so $$n(n+1)$$ is an integer and in fact even. Multiplying by 2 keeps it even, and the extra +1 then forces the whole expression to be odd.

4. Expression D: $$n^3-n+1$$

   First observe that $$n^3-n=n(n^2-1)=n(n-1)(n+1)$$, the product of three consecutive integers. Among any three consecutive integers there is always at least one even factor, so $$n^3-n$$ is even. Adding 1 turns an even number into an odd number, so $$n^3-n+1$$ is always odd.

The reasoning for each case can be summarised in one line:

What is a magic square?
A 3 × 3 magic square is a 3 × 3 grid in which the numbers 1 – 9 are placed so that every row, every column and the two main diagonals add up to the same total. That common total is called the magic constant.

Step 1 – Find the magic constant.
The nine numbers add up to $$1+2+\dots+9 = 45.$$ Since the three rows together cover all nine cells exactly once, each row must total $$45 \div 3 = 15.$$ Hence every row, column and diagonal sums to

Step 2 – The centre must be 5.
The centre cell lies in four lines: the middle row, the middle column and the two diagonals. Together these four lines cover the centre four times and every other cell exactly once, so their total is

$$4 \times (\text{centre}) + (45 - \text{centre}) = 4 \times 15 = 60.$$

This simplifies to $$3 \times (\text{centre}) = 15,$$ giving

Step 3 – Opposite cells add to 10.
Any cell, the centre, and the cell diametrically opposite all lie on one line that sums to 15. Since the centre is 5, the two opposite cells must sum to $$15 - 5 = 10.$$ So the opposite-cell pairs are $$(1,9),\,(2,8),\,(3,7)\text{ and }(4,6).$$

Step 4 – 1 cannot sit in a corner.
Suppose 1 occupied a corner. Then the row, column and diagonal through that corner each need the other two entries to sum to $$15 - 1 = 14.$$

• For the diagonal: the other two entries are the centre 5 and the opposite corner. So the opposite corner is $$14 - 5 = 9.$$
• For the row: two cells (chosen from $$\{2,3,4,6,7,8\}$$, since 5 and 9 are already placed) must add to 14. The only such pair is $$(6,8).$$
• For the column: two more cells from $$\{2,3,4,6,7,8\} \setminus \{6,8\} = \{2,3,4,7\}$$ must also add to 14, but no such pair exists in this remaining set.

That contradiction shows 1 cannot be in a corner. The same argument (or the symmetry $$x \leftrightarrow 10-x$$) rules out 3, 7 and 9 from the corners. So the four odd numbers 1, 3, 7, 9 fill the four edge cells, while the four even numbers 2, 4, 6, 8 fill the four corner cells.

Step 5 – Count the arrangements.

1. Place 1 in any of the 4 edge cells — 4 choices. Once placed, 9 is forced into the opposite edge.
2. Place 3 in one of the remaining 2 edge cells — 2 choices. Once placed, 7 is forced.
3. The four corners now have to hold 2, 4, 6, 8. The row and column sums (each equal to 15) determine every corner uniquely. For example, with 1 on top and 3 on the left, the top row corners must sum to 14 (so they are 6 and 8), the bottom row corners must sum to 6 (so they are 2 and 4), and the left column corners must sum to 12 — only $$8+4$$ works, fixing the top-left as 8 and the bottom-left as 4. The remaining two corners then follow.

Total count $$ = 4 \times 2 \times 1 = 8.$$

Step 6 – The 8 squares are exactly the rotations and reflections of one standard square.
A 3 × 3 grid has exactly 8 symmetries: 4 rotations (0°, 90°, 180°, 270°) and 4 reflections (in the horizontal, vertical and two diagonal mirror lines). Applying any of these to the standard Lo Shu square

produces another magic square, and Step 5 shows there are exactly 8 in all. So every 3 × 3 magic square using 1 – 9 is obtained from this one by a rotation or a reflection.

(If two squares that are rotations or reflections of each other are counted as the same, then there is essentially only one 3 × 3 magic square with the numbers 1 – 9.)

Answer. There are exactly 8 different 3 × 3 magic squares that can be made with the numbers 1 – 9.

Step 1 : Recall the well-known 3 × 3 magic square for the numbers $$1 - 9$$

Every row, column and the two main diagonals add up to $$15$$.

Step 2 : Observe what we need for the new square

• The list $$2,3,4,5,6,7,8,9,10$$ also contains exactly nine consecutive integers.
• If we add the same number to each entry of a magic square, each line still has equal sum, because we have merely increased every term in that line by the same amount.
• Adding $$1$$ to every entry of the square of Step 1 converts $$1\!\to\!2,\,2\!\to\!3,\ldots,9\!\to\!10$$ — precisely the list we want.

Step 3 : Add $$1$$ to each entry

Step 4 : Verify the magic property

Thus $$18$$ is the new magic sum.

Step 5 : Why the method works

• Originally each line was $$15$$. Adding $$1$$ to every cell in that line adds $$3\times1=3$$ to the line–sum.
• Therefore the common line–sum becomes $$15+3=18$$ and the square remains magic.

Step 6 : Comparison with the $$1-9$$ magic square

Strategy summarised: Take the standard $$1-9$$ magic square and add $$1$$ to every entry. Because the same amount is added everywhere, the “magic” property is preserved and we automatically get a correct square for $$2-10$$.

Step 1 – Recall the definition of a magic square

A square array of numbers is called magic if the sum of every row, every column and the two main diagonals is the same. If that common sum is $$S$$, the square is said to have magic constant $$S$$.

Step 2 – Notation for an arbitrary 3 × 3 magic square

Write a general 3 × 3 magic square as

\[\begin{array}{ccc} a & b & c\\ d & e & f\\ g & h & i \end{array}\]

with

$$a+b+c = d+e+f = g+h+i = S$$,
$$a+d+g = b+e+h = c+f+i = S$$,
$$a+e+i = c+e+g = S.$$

Step 3 – Adding the same number to every entry

Choose any constant $$k$$ and form a new square whose entries are $$a+k,\,b+k,\,\ldots,\,i+k$$. A typical row sum becomes

$$(a+k)+(b+k)+(c+k)=a+b+c+3k=S+3k.$$

Exactly the same extra $$3k$$ appears in every row, column and diagonal, so all these sums are still equal. Hence adding (or subtracting) the same constant to every entry keeps the square magic.

Step 4 – Multiplying every entry by the same number

If we multiply each entry by a non-zero constant $$m$$, a typical row sum becomes

$$m(a+b+c)=mS.$$

Every other row, column and diagonal also sums to $$mS$$, so multiplying (or dividing) each entry by the same non-zero constant again gives a magic square.

Step 5 – Rotations and reflections of the whole square

• Rotate the square through $$90^{\circ},\,180^{\circ}$$ or $$270^{\circ}$$; or
• reflect it in the horizontal, vertical, or either diagonal mirror line.

These actions merely re-arrange the nine numbers; they do not change which three numbers lie in any particular row, column or diagonal. Therefore the common sum remains $$S$$ and the new arrangement is also magic.

Step 6 – Swapping the outer rows or outer columns

For a 3 × 3 square, exchanging the first and third rows has the same effect as a vertical reflection, while exchanging the first and third columns is the same as a horizontal reflection. Both have already been proved safe in Step 5, so they also preserve the magic property.

Conclusion

Starting from one magic square we can obtain another magic square by any of the following operations:

• add (or subtract) the same constant to every entry,
• multiply (or divide) every entry by the same non-zero constant,
• rotate the entire square through $$90^{\circ},\,180^{\circ},\,270^{\circ}$$,
• reflect the square in any horizontal, vertical or diagonal mirror line (equivalently, interchange the outer rows or the outer columns).

What we mean by a 3 × 3 magic square
A 3 × 3 arrangement is called a magic square if the three row–sums, the three column–sums and the two diagonal–sums are all the same. For nine consecutive numbers this common value is called the magic sum.

Step 1 : Recall the basic (Lo Shu) magic square for 1 to 9

Every line here adds to $$15$$.

Step 2 : Observe why it works

• The centre is the average of 1 to 9, that is $$5$$.
• Add the same integer $$c$$ to every entry: each line increases by $$3c$$ (because it contains 3 numbers), so the magic property is unchanged.

Step 3 : Replace 1 – 9 by any 9 consecutive numbers

Suppose we want to use the block

$$a,a+1,a+2,\ldots ,a+8$$   (so $$a$$ is the smallest, $$a+8$$ the largest).

This can be done by the shift

$$c = a-1$$

Add $$c$$ to every cell of the Lo Shu square. Symbolically:

The resulting square now contains exactly the required nine numbers, once each.

Step 4 : The new magic sum

Originally each line totalled $$15$$. After adding $$c$$ to each of the three numbers in that line we get

$$\text{new sum}=15+3c=15+3(a-1)=3a+12=3\bigl(a+4\bigr).$$

Notice that $$a+4$$ is the middle (5th) number of the consecutive block, so

Step 5 : Two complete examples

(i) Using 2 – 10 (here $$a=2$$, so $$c=1$$)

Row, column and diagonal sums = $$3(2+4)=18$$.

(ii) Using 9 – 17 (here $$a=9$$, so $$c=8$$)

All line sums = $$3(9+4)=39$$.

Step 6 : A second (algebraic) way – starting from the middle

1. For any 9 consecutive numbers the middle one is $$m=a+4$$. Place it in the centre cell.
2. Copy the pattern of differences from the basic square:
   
   • Corners: $$m+3,m-4,m+1,m-2$$ arranged exactly like 8,1,6,3.
   • Edge-centres: $$m+2,m-1,m+4,m-3$$ arranged like 7,4,9,2.
3. This automatically supplies every number from $$a$$ to $$a+8$$ once each and keeps the magic property.

Conclusion

To make a 3 × 3 magic square from any block of 9 consecutive numbers, simply add the same shift $$a-1$$ to every entry of the standard 1–9 square (or, equivalently, copy its pattern of differences around the new middle number). The magic sum will always be three times the middle number of the chosen block.

Step 1 : Choosing a suitable magic square made with consecutive numbers

The best-known 3 × 3 magic square with the nine consecutive numbers 1 to 9 is

Check that every row, column and the two diagonals add up to 15.

Step 2 : Naming the middle number by a letter

Let the number in the centre be the letter-number $$m$$. For this square the centre actually equals 5, so $$m=5$$, but we shall now describe every entry only in relation to $$m$$.

Step 3 : Writing each number as “how much more or less than” $$m$$

Step 4 : Re-displaying the whole magic square in algebraic form

Replacing every numeral by the expression we just found gives

Step 5 : Verifying that it is still a magic square in algebra

• First row: $$(m+3)+(m-4)+(m+1)=3m$$
• Second row: $$(m-2)+m+(m+2)=3m$$
• Third row: $$(m-1)+(m+4)+(m-3)=3m$$
• Exactly the same check works for every column and the two diagonals, each time giving $$3m$$.

Thus every entry of the square can be described simply by saying how many more or less it is than the centre number $$m$$, and with those expressions the square keeps all the magic-square properties.

Step 1 : Write the two–digit number algebraically
Let the digits be a (tens) and b (ones), where $$1\le a\le 9$$ and $$0\le b\le 9$$.
Hence the number is  $$N = 10a + b$$.

Step 2 : Form the number obtained on interchanging the digits
After reversing the digits the new number is $$M = 10b + a$$.

Step 3 : Add the two numbers

• Sum : $$S = N + M = (10a + b) + (10b + a) = 11(a + b)$$.

Step 4 : Subtract the two numbers
Without loss of generality assume $$a > b$$ (if not, we can simply change the order while subtracting). Then

• Difference : $$D = N - M = (10a + b) - (10b + a) = 9(a - b)$$.

Step 5 : Observations to share with the class

1. The sum of a two–digit number and the number obtained by reversing its digits is always divisible by 11, in fact it equals $$11\times(\text{sum of the digits})$$.
2. The difference of the same two numbers is always divisible by 9, in fact it equals $$9\times(\text{difference of the digits})$$.
3. Since $$11(a+b)$$ and $$9(a-b)$$ are true for all admissible digits, you can verify the above properties quickly for any concrete example chosen in class.

Step 1 : Why the name “Chautīsā Yantra”?
“Chautīs” is the Hindi word for 34. In a magic square the same total, called the magic sum, appears in every row, every column and both main diagonals. Let us check:

Rows
$$7+12+1+14 = 34$$
$$2+13+8+11 = 34$$
$$16+3+10+5 = 34$$
$$9+6+15+4 = 34$$

Columns
$$7+2+16+9 = 34$$
$$12+13+3+6 = 34$$
$$1+8+10+15 = 34$$
$$14+11+5+4 = 34$$

Diagonals
$$7+13+10+4 = 34$$
$$14+8+3+9 = 34$$

Because the common total is 34, the square is called the “Chautīsā (34) Yantra”.

Step 2 : Other groups of four numbers that also make 34

(i) The four corners of the whole square.

$$7+14+9+4 = 34$$

(ii) The four edge-centres (middle numbers of the outer rows and columns).

$$12+1+15+6 = 34$$

(iii) Every 2 × 2 sub-square.

(iv) Broken (wrap-around) diagonals. Pick a starting cell and step one to the right and one down repeatedly, jumping to the opposite side whenever you would leave the square. Starting at the second cell of the top row:

$$12\,(r1c2) + 8\,(r2c3) + 5\,(r3c4) + 9\,(r4c1) = 34$$

Conclusion
Not only the rows, columns and main diagonals, but several other sets of four numbers — the four corners, the four edge-centres, every 2 × 2 block and certain broken diagonals — also add up to 34. That is what makes the Chautīsā Yantra so special.

The 3 × 3 magic square that was derived earlier can be written in the general form

\[\begin{array}{ccc} a+3 & a-4 & a+1 \\ a-2 & a & a+2 \\ a-1 & a+4 & a-3 \end{array}\]

Here the number written in the middle is $$a$$. We are told that the middle (centre) number is 25, so

$$a = 25.$$

Substituting $$a = 25$$ in each of the nine expressions:

• Top–left cell: $$a+3 = 25+3 = 28$$
• Top–middle cell: $$a-4 = 25-4 = 21$$
• Top–right cell: $$a+1 = 25+1 = 26$$
• Middle–left cell: $$a-2 = 25-2 = 23$$
• Centre cell: $$a = 25$$
• Middle–right cell: $$a+2 = 25+2 = 27$$
• Bottom–left cell: $$a-1 = 25-1 = 24$$
• Bottom–middle cell: $$a+4 = 25+4 = 29$$
• Bottom–right cell: $$a-3 = 25-3 = 22$$

Putting these nine numbers back into the grid gives

Verification. Let us check that every row, every column and both main diagonals add up to the same total.

• Rows: $$28+21+26 = 75,\ \ 23+25+27 = 75,\ \ 24+29+22 = 75.$$
• Columns: $$28+23+24 = 75,\ \ 21+25+29 = 75,\ \ 26+27+22 = 75.$$
• Diagonals: $$28+25+22 = 75,\ \ 26+25+24 = 75.$$

All eight line-sums equal 75, which equals $$3 \times 25 = 3a$$, as expected. Hence the table above is the required magic square whose centre entry is 25.

Let the three terms appearing in a row (the same reasoning works for a column or a diagonal) be

$$a - x, \; a, \; a + x$$

for some number $$x$$. (In the magic square given in the textbook each row, column or diagonal contains one term that is less than $$a$$ by the same amount $$x$$ and another that is greater than $$a$$ by the same amount $$x$$; the middle term itself is $$a$$.)

Now add the three terms:

\[ (a - x) + a + (a + x) \]

Combine like terms, keeping the $$x$$–terms and the $$a$$–terms separately:

$$a - x + a + a + x$$

The numbers $$-x$$ and $$+x$$ cancel each other:

$$a + a + a = 3a$$

Hence, whether we choose a row, a column or a diagonal, the sum of the three entries is always

\[ 3a. \]

Goal. Construct a magic square in which every row, column and each of the two main diagonals add up to 60 (this common total is called the magic sum).

Step 1 – Start from a known 3 × 3 magic square.

The famous Lo Shu magic square uses the integers 1 to 9. Each line in it adds up to 15.

Indeed, for the first row:

$$8 + 1 + 6 = 15$$

All other rows, columns and diagonals can be checked in the same way, each giving 15.

Step 2 – Decide how to obtain the required magic sum 60.

The present magic sum is 15. We need 60. Because

$$60 = 4 \times 15,$$

multiplying every entry of the Lo Shu square by 4 will multiply every line’s sum by 4 as well, giving 60.

Step 3 – Multiply every entry by 4.

So the new square is

Step 4 – Verify the magic sum.

• Row 1: $$32 + 4 + 24 = 60$$
• Row 2: $$12 + 20 + 28 = 60$$
• Row 3: $$16 + 36 + 8 = 60$$
• Column 1: $$32 + 12 + 16 = 60$$
• Column 2: $$4 + 20 + 36 = 60$$
• Column 3: $$24 + 28 + 8 = 60$$
• Main diagonal (top-left to bottom-right): $$32 + 20 + 8 = 60$$
• Other diagonal (top-right to bottom-left): $$24 + 20 + 16 = 60$$

Every line gives the required total of 60, so the square truly is magic with the desired sum.

Conclusion. One possible 3 × 3 magic square whose magic sum is 60 is

Step 1 : Recall what a 3 × 3 magic square is
In a 3 × 3 magic square the nine different numbers are arranged so that every row, every column and the two long diagonals have the same total. This common total is called the magic sum.

Step 2 : Start from the well-known Lo-Shu square
The usual (Lo-Shu) magic square uses the consecutive numbers 1 to 9.

\[\begin{array}{ccc} 2 & 7 & 6\\ 9 & 5 & 1\\ 4 & 3 & 8 \end{array}\] Every line adds to $$15$$.

Step 3 : How to get new numbers while keeping the magic property
Suppose every entry $$k$$ of the above square is replaced by a new number of the form $$k' = a + (k-1)d,$$ where $$a$$ and $$d$$ are any integers and $$d \neq 0$$. (When $$d = 1$$ and $$a = 1$$ we go back to 1–9.)

Why does this work?
In one row of the Lo-Shu square the three original positions are, say, $$k_1, k_2, k_3$$. Their sum is $$k_1 + k_2 + k_3 = 15$$. After the change the new row sum is

$$k_1' + k_2' + k_3' = \bigl[a + (k_1-1)d\bigr] + \bigl[a + (k_2-1)d\bigr] + \bigl[a + (k_3-1)d\bigr]$$
$$\;= 3a + \bigl[(k_1+k_2+k_3)-3\bigr]d$$
$$\;= 3a + (15-3)d = 3\bigl(a + 4d\bigr).$$

The same calculation holds for every row, column and diagonal, so the new grid is still a magic square with magic sum \[3\bigl(a + 4d\bigr).\]

Step 4 : Choose non-consecutive numbers and exhibit the square
Take for example $$a = 2,\; d = 2$$. The nine numbers produced are

$$2,\;4,\;6,\;8,\;10,\;12,\;14,\;16,\;18$$

These are clearly not consecutive integers. Put them in the same pattern as the Lo-Shu square:

\[\begin{array}{ccc} 4 & 14 & 12\\ 18 & 10 & 2\\ 8 & 6 & 16 \end{array}\]

Check one line to be sure:
Top row $$4 + 14 + 12 = 30$$;
first column $$4 + 18 + 8 = 30$$;
main diagonal $$4 + 10 + 16 = 30$$. All other lines also add to $$30$$.

Step 5 : General conclusion
Because we can choose any integers $$a$$ and $$d\,(\neq 0)$$, the nine numbers need not be consecutive. As long as we put them according to the above rule, a magic square is always obtained.

Therefore it is possible to make a 3 × 3 magic square with nine non-consecutive numbers.

Understanding the problem
We have to clap out 6 beats using only short beats (value $$1$$) and long beats (value $$2$$).
Every possible ordered sum of $$1$$’s and $$2$$’s whose total is $$6$$ is called a 6-beat rhythm.

Step 1 : Decide how many long beats (2’s) we use
Let $$m$$ be the number of 2’s.

• Because each 2 contributes $$2$$ beats, the largest $$m$$ possible is the whole-number part of $$6\div 2 = 3$$. Thus $$m=0,1,2,3$$.

Step 2 : For each $$m$$ find the remaining number of short beats (1’s)
If there are $$m$$ twos, their total length is $$2m$$, leaving $$6-2m$$ beats to be filled with 1’s.

Step 3 : Arrange the beats in a row
The row length (number of symbols) for each $$m$$ is
$$\bigl(m\text{ twos}\bigr) + \bigl(6-2m\text{ ones}\bigr)= 6-m.$$ The systematic way is: “Choose positions for the twos; the remaining places automatically hold ones.”

Step 4 : Total count

So there are 13 distinct 6-beat rhythms.

Given sequence   $$1,\;2,\;3,\;5,\;8,\;13,\;21,\;34,\;55,\;89,\;\_\_\,\;\_\_\,\;\_\_\,\;\ldots$$

Look at how each term is obtained:

• $$1+2=3$$
• $$2+3=5$$
• $$3+5=8$$
• and\;so\;on\;\ldots$$

So every term (from the third term onward) is the sum of the two just before it. This is called the Fibonacci rule.

Step 1. Find the next three numbers

1. 11^th term   $$=55+89=144$$
2. 12^th term   $$=89+144=233$$
3. 13^th term   $$=144+233=377$$

Thus the sequence now reads
$$\ldots,55,\;89,\;144,\;233,\;377,\;\ldots$$

Step 2. Will the next (14^th) term be odd or even ?

Instead of adding again, notice the odd–even pattern that the rule produces.

Reading the last column we get the repeating pattern

$$O,\;E,\;O,\;O,\;E,\;O,\;O,\;E,\;O,\;\ldots$$

That is O E O keeps repeating every three places.

The 13^th term is the second O in a fresh block “O E O”.
Therefore the 14^th term will be the E of that block – hence it will be even.

(If we did add the numbers, we would get $$377+233=610$$, and indeed 610 is even.)

Step 1 – Recall the definition of the Virahāṅka numbers
The Virahāṅka sequence is the same as the Fibonacci sequence that begins with two 1’s :

• First term : $$F_1 = 1$$
• Second term : $$F_2 = 1$$
• Every later term : $$F_n = F_{n-1} + F_{n-2}\;\;(n \ge 3)$$

Step 2 – Write the first few Virahāṅka numbers and mark each as odd (O) or even (E)

The observed parity string is
$$\text{O, O, E, O, O, E, O, O, E,}\;\ldots$$

Step 3 – Learn how parity behaves under addition

• $$\text{odd} + \text{odd} = \text{even}$$
• $$\text{odd} + \text{even} = \text{odd}$$ (and vice-versa)
• $$\text{even} + \text{even} = \text{even}$$

Step 4 – Prove the pattern “O, O, E” repeats forever

Look at three consecutive Virahāṅka numbers : $$F_{k-2},\;F_{k-1},\;F_k$$.

1. Suppose the first two have parities O and O. Then by the rule above,
     $$F_k = F_{k-1} + F_{k-2} \;\text{is} \; \text{even (E)}.$$
     So the triple is O, O, E.
2. Next triple : $$F_{k-1},\;F_k,\;F_{k+1}$$ has parities O, E, ?. Since $$F_{k+1}=F_k+F_{k-1}$$ is E + O, it is odd (O). Thus we get O, E, O.
3. Following triple : E, O, ?. Again $$F_{k+2}=F_{k+1}+F_k$$ is O + E = O, giving E, O, O.

We have cycled back to two O’s followed by an E. Hence the parities repeat every three terms:

\[\text{Odd, Odd, Even, Odd, Odd, Even, }\ldots\]

Step 5 – State the final result
The parity of the nth Virahāṅka number is

• odd when n leaves a remainder 1 on division by 3,
• odd when n leaves a remainder 2 on division by 3,
• even when n is a multiple of 3.

This matches the repeating pattern : O, O, E, O, O, E, …

Step 1 : Translate the puzzle into an equation

The addition shown is

(one addend)  $$T$$
(second addend) + $$T$$
(third addend) + $$T$$

This says “three times the digit $$T$$ equals the two–digit number whose tens digit is $$U$$ and whose units digit is again $$T$$”. If we let the actual numerical values of the letters be lower-case $$t$$ and $$u$$, the statement becomes

$$3t = 10u + t$$    (because $$UT$$ means $$10u + t$$).

Step 2 : Solve the equation

Bring like terms together:

$$3t - t = 10u \Rightarrow 2t = 10u.$$ Divide by 2:

$$t = 5u.$$

Step 3 : Use the fact that both $$t$$ and $$u$$ are digits (0 – 9)

• If $$u = 0$$, then $$t = 0$$, but the sum would not be a two-digit number, so discard.
• If $$u = 1$$, then $$t = 5$$  (allowed, both are single digits).
• If $$u \ge 2$$, then $$t = 5u \ge 10$$, which is not a single digit. Hence impossible.

Therefore the only possible assignment is

$$U = 1, \; T = 5.$$

Step 4 : Answer the extra questions

• Can $$T$$ be 2?
If $$T = 2$$, the left side is $$2 + 2 + 2 = 6$$, a one-digit number, so it cannot be written as $$UT$$ with two digits and units digit 2. Hence no.

• Can $$T$$ be 3?
If $$T = 3$$, the left side is $$3 + 3 + 3 = 9$$, again only one digit, so no.

Conclusion

The unique solution consistent with all conditions is

Step 1 : Change the verbal statement into an equation
The two–digit number $$K2$$ can be written in numerals as $$10K+2$$.
Adding it to itself gives
\[ (10K+2)+(10K+2)=20K+4. \]
The result has been described as the three–digit number $$\text{HMM}$$, so in numerals
\[ \text{HMM}=100H+10M+M=100H+11M. \]

Step 2 : The units digit fixes $$M$$
The units digit of $$20K+4$$ is clearly $$4$$, and in $$\text{HMM}$$ the units digit is $$M$$. Therefore
$$M=4.$$
So the sum must look like $$\text{H44}$$.

Step 3 : Work on the tens column to find $$K$$
When the two numbers are added in columns we have

• No carry to the hundreds place: Then $$2K<10$$ and the tens digit is exactly $$2K$$, which must be $$4$$.
   $$2K=4\;\Rightarrow\;K=2.$$
   But $$K=2$$ would make the addends $22+22=44$, a two-digit answer, contradicting the fact that the sum is three-digit. Hence this case is impossible.
• Carry 1 to the hundreds place: Then $$2K\ge 10$$. The tens digit left after subtracting the carried 10 is $$2K-10$$. This digit must again be $$4$$.
   $$2K-10=4\;\Rightarrow\;2K=14\;\Rightarrow\;K=7.$$
   Because $$K=7$$ really is a digit and $72+72=144$ is three-digit, this case works.

Step 4 : Identify $$H$$
With $$K=7$$ the addition is

The hundreds digit is therefore $$H=1$$.

Step 5 : Answer the questions
$$H$$ comes out to be $$1$$ only. It is not $$2$$, and it is not $$3$$ either.

The bulb starts in the ON position.

Pressing (toggling) the switch once changes the state of the bulb. Therefore each press reverses whatever state the bulb is in at that moment.

Let us look at the first few presses to see the pattern:

We observe:

• After every even number of presses (2, 4, 6, …) the bulb is again ON.
• After every odd number of presses (1, 3, 5, …) the bulb is OFF.

Mathematically, an even number can be written as $$2k$$ and an odd number as $$2k + 1$$, where $$k$$ is a whole number.

Dorjee presses the switch $$77$$ times. We check whether $$77$$ is odd or even:

\[77 = 2 \times 38 + 1\]

Since there is a remainder of $$1$$, $$77$$ is an odd number of presses.

Because an odd number of presses leaves the bulb in the opposite state to the one it began with, and it began ON, the bulb must now be OFF.

Therefore, after 77 toggles, the light bulb will be OFF.

Number the pages of a book in the usual way: 1, 2, 3, 4 … ; every odd-numbered page is on the front of a sheet and the next even page is on its back.

Take one loose sheet. If the smaller page number on it is $$n$$, then the two page numbers are

$$n \text{ (front)}, \; n+1 \text{ (back)}$$

Because the front page is always odd, $$n=2k+1$$ for some whole number $$k$$.

Sum on that sheet:

$$n+(n+1)=(2k+1)+(2k+2)=4k+3$$

When any number of the form $$4k+3$$ is divided by 4 the remainder is 3. (We write this as “$$4k+3\equiv3\;(\text{mod}\,4)$$”.)

Step 1 Remainder given by one sheet

Each single sheet therefore contributes a sum that leaves remainder 3 on division by 4.

Step 2 Remainder given by 50 sheets

With 50 such sheets the total remainder is

$$50\times3=150$$

Dividing 150 by 4:

$$150=4\times37+2$$

So the total of the page numbers of 50 sheets must leave remainder 2 when divided by 4, i.e.

Step 3 Check the proposed total 6000

Now divide 6000 by 4:

$$6000=4\times1500+0$$

It leaves remainder 0, not 2.

Conclusion

Since the sum of the page numbers of any 50 sheets must give remainder 2 on division by 4, while 6000 gives remainder 0, the number 6000 can never be obtained in this situation.

Therefore, Liswini’s loose pages cannot have a total of 6000.

Step 1 – Name the six boxes

Let the entries be

We only care whether each letter is odd (write 1) or even (write 0).

Step 2 – Translate the given parities

• Row 1 is odd  ⇒  $$a\oplus b\oplus c = 1$$
• Row 2 is even  ⇒  $$d\oplus e\oplus f = 0$$
• Column 1 is even  ⇒  $$a\oplus d = 0$$
• Column 2 is even  ⇒  $$b\oplus e = 0$$
• Column 3 is odd  ⇒  $$c\oplus f = 1$$

(Here $$\oplus$$ means “add modulo 2”: 0 for even, 1 for odd.)

Step 3 – Express everything through $$a,\,b,\,c$$

From the column relations:

• $$d = a$$  (same parity as $$a$$)
• $$e = b$$  (same parity as $$b$$)
• $$f = 1 \oplus c$$  (opposite parity to $$c$$)

Step 4 – Check that Row 2 automatically works

Substitute the above in the Row 2 condition:

so Row 2 is indeed even. No extra restriction appears.

Step 5 – See how many odds we get

There must be exactly three odds and three evens.

The number of odds is

(because the pair $$c,f$$ always contributes exactly one odd).

We want this equal to 3, hence

So exactly one of $$a$$ and $$b$$ is odd and the other is even. Take two cases.

Step 6 – Choose actual numbers

Case 1 (take $$a$$ odd, $$b$$ even):

• $$a = 1$$ (odd)
• $$b = 2$$ (even)
• Row 1 must be odd, so $$c$$ must be even. Take $$c = 4$$.
• Then $$d = a =$$ an odd number, choose $$d = 3$$.
• $$e = b =$$ an even number, choose $$e = 6$$.
• $$f$$ has opposite parity to $$c$$, so $$f$$ must be odd. Take $$f = 5$$.

Step 7 – Write the completed grid

Check:

• Row 1 sum = 1 + 2 + 4 = 7 (odd)
• Row 2 sum = 3 + 6 + 5 = 14 (even)
• Column 1 sum = 1 + 3 = 4 (even)
• Column 2 sum = 2 + 6 = 8 (even)
• Column 3 sum = 4 + 5 = 9 (odd)

All required parities are satisfied and we have used exactly three odd numbers (1, 3, 5) and three even numbers (2, 4, 6).

Note: The second case ($$a$$ even, $$b$$ odd) also gives correct solutions; one such is

Either arrangement earns full credit.

Step 1 – Recall a standard $$3 \times 3$$ magic square

In this square every row, column and diagonal totals $$15$$.

Step 2 – Make the magic sum zero

• We want each line to add to $$0$$, not $$15$$.
• If we subtract the same number $$k$$ from every entry, each line loses $$3k$$ (because it has three entries).
• Choose $$k = 5$$ so that $$15 - 3\times 5 = 0$$.

Subtracting $$5$$ from every entry:

Step 3 – Write the new square clearly

Step 4 – Verify the magic property

• Rows: $$3+(-4)+1 = 0$$, $$-2+0+2 = 0$$, $$-1+4+(-3) = 0$$.
• Columns: $$3+(-2)+(-1) = 0$$, $$-4+0+4 = 0$$, $$1+2+(-3) = 0$$.
• Diagonals: $$3+0+(-3) = 0$$ and $$1+0+(-1) = 0$$.

Every line sums to $$0$$, and not all entries are zero, so the requirement is satisfied.

Step 1 : Understanding “parity”
A whole number is called even if it is divisible by $$2$$ and odd otherwise. So, to know the parity of any number we only need to decide whether it is a multiple of $$2$$.

Step 2 : Writing the required sum
The question asks for the parity of the sum of all natural numbers from $$1$$ to $$100$$, that is
$$1 + 2 + 3 + \dots + 98 + 99 + 100$$.

Step 3 : Adding the numbers (pairing method)
A quick way for a Class 7 student is to pair the first and last terms:

• First pair: $$1 + 100 = 101$$
• Second pair: $$2 + 99 = 101$$
• Third pair: $$3 + 98 = 101$$
• \(\dots\)
• Fiftieth pair: $$50 + 51 = 101$$

There are $$50$$ such pairs and each pair gives $$101$$. Therefore the total is

Check with the formula  $$\displaystyle \frac{n(n+1)}{2}$$ for the sum of the first $$n$$ natural numbers:

Put $$n = 100$$:

$$\frac{100(100+1)}{2} = \frac{100\times101}{2} = 50\times101 = 5050$$  ✓

Step 4 : Deciding the parity of 5050
The last digit of $$5050$$ is $$0$$, which shows it is divisible by $$2$$. Formally:

$$5050 \div 2 = 2525$$ with no remainder, so $$5050$$ is even.

Conclusion
The sum of all numbers from $$1$$ to $$100$$ is $$5050$$, an even number. Hence the required sum has even parity.

The Virahāṅka sequence is the same as the well-known Fibonacci sequence. Each term is the sum of the two terms that come just before it.

Let the two given consecutive terms be the n-th and (n+1)-th terms:

$$F_n = 987,\;F_{n+1} = 1597.$$

Step 1: Find the next two numbers.

• Next term:
  $$F_{n+2} = F_{n+1} + F_n = 1597 + 987 = 2584.$$
• Term after that:
  $$F_{n+3} = F_{n+2} + F_{n+1} = 2584 + 1597 = 4181.$$

Step 2: Find the previous two numbers.

To go backward in a Fibonacci (Virahāṅka) sequence, subtract the smaller term of a consecutive pair from the larger:

• Immediate previous term:
  $$F_{n-1} = F_{n+1} - F_n = 1597 - 987 = 610.$$
• One more term back:
  $$F_{n-2} = F_n - F_{n-1} = 987 - 610 = 377.$$

Check: $$377 + 610 = 987$$ and $$610 + 987 = 1597$$, confirming the values.

Hence the required portion of the Virahāṅka sequence, written in increasing order, is

\[377,\;610,\;987,\;1597,\;2584,\;4181.\]

What is counted? Each path is a sequence that adds up to 8 using only the numbers 1 (one-step) and 2 (two-step). The order in the sequence matters.

Method 1 — Building a recurrence ("add the last step")

Let $$f(n)$$ be the number of different ways to climb $$n$$ steps.

If Angaan is already on step $$n-1$$, one more single step takes him to $$n$$. This contributes $$f(n-1)$$ possibilities.
If Angaan is on step $$n-2$$, one more double step (2 at a time) takes him to $$n$$. This contributes $$f(n-2)$$ possibilities.

So every way to reach step $$n$$ is obtained by appending the last move to a way of reaching either $$n-1$$ or $$n-2$$.

Therefore

$$f(n)=f(n-1)+f(n-2).$$

Starting values.
$$f(1)=1$$ (only 1) and $$f(2)=2$$ (1,1 or 2).

Compute successively up to $$n=8$$.

The required number is

\[34\]

Method 2 — Counting by how many 2-steps are used (an optional check)

Suppose the path contains $$k$$ jumps of 2 steps. Then $$k$$ can be $$0,1,2,3,4$$ (because $$4\times2=8$$ is the maximum).

With $$k$$ two-steps, the remaining $$8-2k$$ jumps are one-steps. Altogether there are $$k+(8-2k)=8-k$$ moves to arrange. The number of ways to decide which of these $$8-k$$ positions are the two-steps is the combination $$\binom{8-k}{k}$$.

Adding these: $$1+7+15+10+1=34$$, confirming the previous result.

Hence, Angaan can climb the 8-step staircase in 34 different ways.

Step 1 – Recall the definition of the Virahāṅka sequence

The Virahāṅka numbers are exactly the Fibonacci numbers, generated by the rule

$$T_1 = 1, \; T_2 = 1, \; \text{and for every } n \ge 3,\; T_n = T_{n-1}+T_{n-2}.$$

Thus the list begins

$$1,\;1,\;2,\;3,\;5,\;8,\;13,\;21,\;34,\;55,\;\ldots$$

Step 2 – Set up a quick way to read the parity (even / odd)

It is enough to look at each term modulo 2. Write

$$E \text{ for an even number (remainder }0\text{ mod }2), \quad O \text{ for an odd number (remainder }1\text{ mod }2).$$

Addition rule for parity:

$$O+O = E,\;\; O+E = O,\;\; E+O = O,\;\; E+E = E.$$

Step 3 – Work out the parity pattern once and for all

Observe the repeating block

$$O,\;O,\;E \quad\;(\text{length }3).$$

To be sure it really repeats, notice that if two consecutive parities repeat, the next one is forced:

• Whenever we again meet the pair $$O,O,$$ their sum is $$E,$$ restarting the cycle.

Therefore the parities follow the fixed 3-term cycle

$$O,\;O,\;E,\;O,\;O,\;E,\;O,\;O,\;E,\;\ldots$$

Step 4 – Locate the 20^th term in the cycle

Because the cycle length is $$3,$$ divide the position number by $$3$$:

$$20 \div 3 = 6 \text{ remainder } 2.$$

The remainder tells us which place inside the block the term occupies:

• remainder 1 → first entry → $$O$$
• remainder 2 → second entry → $$O$$
• remainder 0 → third entry → $$E$$

So $$T_{20}$$ is odd.

Step 5 – Check by direct computation (optional but reassuring)

Continuing the list quickly:

Indeed $$6765$$ is odd, confirming the rule-based answer.

Conclusion

The 20^th Virahāṅka (Fibonacci) number is odd.

Step 1 : Set up column-wise addition

(The sum also has a hundreds digit  $$T$$.)

Step 2 : Work in the ones column

The ones column says

$$T + A = T + 10\times c_1$$

where $$c_1$$ is the carry sent to the tens column.

Subtract $$T$$ from both sides:  $$A = 10c_1$$.

• If $$c_1 = 0$$  ⇒  $$A = 0$$ (allowed).
• If $$c_1 = 1$$  ⇒  $$A = 10$$ (impossible for a single digit).

Hence $$c_1 = 0$$ and therefore

$$A = 0$$.

Step 3 : Work in the tens column

Now

$$U + T + c_1 = A + 10c_2.$$ Because $$c_1 = 0$$ and $$A = 0$$, this becomes

$$U + T = 10c_2.$$

The largest possible value of $$U + T$$ is $$9 + 8 = 17 < 20$$, so the only possible multiple of 10 is 10.

Therefore

$$U + T = 10 \\ c_2 = 1.$$

Step 4 : Work in the hundreds column

The hundreds digit of the sum is $$T$$, and it is produced solely by the carry from the tens column, so

$$c_2 = T.$$

But Step 3 gave $$c_2 = 1$$, hence

$$T = 1.$$

Step 5 : Find $$U$$

Use $$U + T = 10$$:

$$U + 1 = 10 \implies U = 9.$$

Step 6 : Write the numbers and check

$$UT = 91, \; TA = 10, \; TAT = 101.$$

Check:  $$91 + 10 = 101$$ — correct, and all letters have distinct digits.

Thus the only solution is

$$U = 9, \; T = 1, \; A = 0.$$