Chapter 1: Large Numbers Around Us

Step 1 · Convert the word “lakh” into a number
One lakh = $$1\,00\,000$$ varieties of rice.

Step 2 · How many days are there in 100 years?
A year is taken as $$365$$ days in ordinary calculations for Class 7.
Therefore, in $$100$$ years the number of days is
$$100\times 365 = 36\,500$$ days.

(If we include leap years, there will be about 25 extra days, making it $$36\,525$$ days. The conclusion does not change.)

Step 3 · How many varieties can be tasted in that time?
Trying one new variety each day for $$100$$ years gives
$$36\,500$$ varieties.

Step 4 · Compare with the total number of varieties
Required varieties = $$1\,00\,000$$
Tasted varieties = $$36\,500$$

Difference = $$1\,00\,000 - 36\,500 = 63\,500$$ varieties still left untasted.

Step 5 · Check how close we get
The fraction of varieties tasted is $$\dfrac{36\,500}{1\,00\,000}=0.365=36.5\%$$.

Conclusion
Even in a full life of $$100$$ years, Estu would taste only about one-third of all the rice varieties. So he would not come close to tasting every kind.

We move from one “family” of numbers to the next simply by adding 1 to the present largest number.

• Largest 3-digit number:
  The digits in the hundreds, tens and ones places are all 9, so the greatest 3-digit number is
  $$999$$.
• Smallest 4-digit number:
  Add 1 to the largest 3-digit number:
  $$999 + 1 = 1000$$.
  So the first (smallest) 4-digit number is $$1000$$.
• Largest 4-digit number:
  Put 9 in every place up to the thousands place:
  $$9999$$.
• Smallest 5-digit number:
  Add 1 to the largest 4-digit number:
  $$9999 + 1 = 10000$$.
  Therefore the first 5-digit number is $$10000$$.
• Largest 5-digit number:
  Again, write 9 in every place up to the ten-thousands place:
  $$99999$$.
• Smallest 6-digit number:
  Add 1 to the largest 5-digit number:
  $$99999 + 1 = 100000$$, which we read as “One Lakh”.

The portion given is

$$99,995 \rightarrow 99,996 \rightarrow \boxed{\phantom{0}} \rightarrow 99,998 \rightarrow \boxed{\phantom{0}} \rightarrow \boxed{\phantom{0}} \rightarrow \boxed{\phantom{0}} \rightarrow \boxed{\phantom{0}} \rightarrow \boxed{\phantom{0}}$$

1. Detect the pattern

Difference of the first two terms:

$$99,996-99,995 = 1$$

So each successive term is obtained by adding 1.

2. Fill the blanks successively

• 3rd term: $$99,996 + 1 = 99,997$$
• 4th term (given) is $$99,998$$  ✔️
• 5th term: $$99,998 + 1 = 99,999$$
• 6th term: $$99,999 + 1 = 100,000$$
• 7th term: $$100,000 + 1 = 100,001$$
• 8th term: $$100,001 + 1 = 100,002$$
• 9th term: $$100,002 + 1 = 100,003$$

3. Completed sequence

\[99,995 \rightarrow 99,996 \rightarrow 99,997 \rightarrow 99,998 \rightarrow 99,999 \rightarrow 100,000 \rightarrow 100,001 \rightarrow 100,002 \rightarrow 100,003\]

Given
Total varieties to be tasted  =  $$1\text{ lakh}=1\,00\,000$$.

1. Convert 100 years into days
One calendar year is taken as $$365$$ days at this level, so

$$100\text{ years}=100\times 365=36\,500\text{ days}.$$

2. If we eat 2 varieties per day

• Varieties that can be tasted in the whole lifetime:
  $$36\,500\times 2 = 73\,000.$$
• Since $$73\,000 < 1\,00\,000$$, 2 per day is not enough.

Equivalently, to finish all the varieties:

$$\frac{1\,00\,000}{2}=50\,000\text{ days},\qquad \frac{50\,000}{365}\approx 137\text{ years} > 100.$$

3. If we eat 3 varieties per day

• Varieties that can be tasted in $$36\,500$$ days:
  $$36\,500\times 3 = 1\,09\,500.$$
• Because $$1\,09\,500 \ge 1\,00\,000$$, the target can be reached within 100 years.

To see how many years are actually needed:

$$\frac{1\,00\,000}{3}=33\,333.\overline{3}\text{ days},$$
$$\frac{33\,333.\overline{3}}{365}\approx 91.32\text{ years}.$$

4. Conclusion
Eating 2 varieties per day will not let someone finish all the 1 lakh varieties in 100 years, but eating 3 varieties per day will; it would take only about 91 years.

Step 1 – Estimate a convenient value of $$y$$.
To be near one lakh ( $$1\,00\,000$$ ) days, divide by the number of days in one year:
$$\dfrac{1\,00\,000}{365}\approx 273.97$$.
The closest whole number of years is therefore $$y = 274$$.

Step 2 – Find the exact number of days in 274 years.
$$\text{Days}=365\times274$$

• First multiply by 200: $$365\times200 = 73\,000$$
• Then by 70: $$365\times70 = 25\,550$$
• Then by 4: $$365\times4 = 1\,460$$

Add the three partial products:
$$73\,000 + 25\,550 + 1\,460 = 1\,00\,010$$

Step 3 – Compare with one lakh.
One lakh days = $$1\,00\,000$$.
Difference = $$1\,00\,010 - 1\,00\,000 = 10$$.

Conclusion.
If a person lives for $$274$$ years, the calculated number of days in that lifetime (ignoring leap years) is $$1\,00\,010$$, which is only 10 days more than one lakh.

Step 1 – Height of one floor
Somu’s height is $$1\,\text{m}$$. Each floor is about four times Somu’s height, so
$$\text{height of one floor}=4\times1\,\text{m}=4\,\text{m}$$

Step 2 – Number of floors
Counting the floors in the picture we get $$10$$ floors.

Step 3 – Approximate height of building
$$\text{height of building}=\text{number of floors}\times\text{height of one floor}$$
$$=10\times4\,\text{m}=40\,\text{m}$$

So, the building is approximately

\[\boxed{40\,\text{metres}}\]

Step 1 : Write down the given heights

• Height of the Statue of Unity   = $$182\;\text{m}$$
• Height of the building (given in the textbook picture)   = $$73\;\text{m}$$

Step 2 : Compare the two heights

The Statue of Unity is written with three digits (hundreds–tens–ones), while the building is written with only two digits (tens–ones). Therefore, without any calculation we can already say that

$$182 > 73$$

Hence, the Statue of Unity is taller.

Step 3 : Find how much taller

We subtract the shorter height from the taller one.

Working it out step by step:

• Ones: $$2-3$$ is not possible, so we borrow $$1$$ ten (which is $$10$$ ones) from the tens column.
    New calculation: $$(2+10)-3 = 12-3 = 9$$ ones.
• Tens: After borrowing, the tens digit is now $$7$$. So $$7-7 = 0$$ tens.
• Hundreds: $$1-0 = 1$$ hundred.

Thus, the difference is $$109\;\text{m}$$.

Conclusion

The Statue of Unity is taller than the building, and it is $$109\;\text{metres}$$ taller.

Step 1 : Identify the two heights given in the question.
Height of the Kunchikal waterfall  = $$455\text{ m}$$
Height of Somu's building  = $$13\text{ m}$$

Step 2 : Write the subtraction needed.
To find how much taller the waterfall is, subtract the building’s height from the waterfall’s height:
$$455\text{ m}-13\text{ m}$$

Step 3 : Do the subtraction neatly.

Start from the ones place:

• Ones : $$5-3=2$$
• Tens : $$5-1=4$$
• Hundreds : $$4-0=4$$

So the difference is $$442\text{ m}$$.

Step 4 : Conclude.
The Kunchikal waterfall is $$442\text{ m}$$ taller than Somu’s building.

Step 1 – Note the heights given in the question.

• Height of the waterfall (from the textbook picture of Kunchikal) = $$450\,\text{m}$$.
• Somu is $$1\,\text{m}$$ tall, and each floor of his building is four times Somu's height, so the height of one floor = $$4 \times 1\,\text{m} = 4\,\text{m}$$.

Step 2 – Set up the division.

The number of floors needed equals the height of the waterfall divided by the height of one floor:

Step 3 – Carry out the division.

• The largest multiple of $$4$$ that does not exceed $$450$$ is $$4 \times 112 = 448$$.
• The remainder is $$450 - 448 = 2$$, which is less than $$4$$.
• So $$\dfrac{450}{4} = 112\dfrac{2}{4} = 112\dfrac{1}{2} = 112.5$$.

This means $$112$$ floors give a height of only $$448\,\text{m}$$, falling $$2\,\text{m}$$ short of the waterfall.

Step 4 – Round up to a whole floor.

Half a floor cannot be built, and the building must be at least as tall as the waterfall, so we round $$112.5$$ up to the next whole number.

Therefore, Somu's building should have $$113$$ floors.

Step 1 – Write the number called “a lakh”

In the Indian system we put commas after every two digits from the right (except the first group of three). So a lakh is written as

$$1,00,000$$

This is the same as

$$1\times10^5=100\,000$$

Step 2 – Compare it with quantities a Class 7 student already knows

• Money in the pocket. If your daily pocket money is Rs 20, then in one year (365 days) you collect $$20\times365=7\,300\text{ rupees}$$ To reach one lakh rupees you would need $$\frac{1,00,000}{7,300}\approx13.7\text{ years}$$ For personal spending, one lakh therefore feels big.
• Number of seconds. 1 lakh seconds is $$100\,000\text{ s}=\frac{100\,000}{3,600}\text{ h}\approx27.8\text{ h}$$ —that is only a little more than one day, so here a lakh feels small.
• India’s population. India has about 138 crore people. $$138\text{ crore}=1,38,00,00,000$$ A lakh people are only $$\frac{1,00,000}{1,38,00,00,000}\times100\%\approx0.0072\,\%$$ of the total. In a national census, one lakh is again quite small.

Step 3 – State the conclusion clearly

Whether a lakh is “big” or “small” is relative to what you are comparing it with:

• For everyday money, a lakh is a large amount.
• Against very large counts such as national budgets, distances in astronomy or India’s population, a lakh is tiny.

So the correct way to “view” a lakh is: it is neither inherently big nor inherently small; its size makes sense only in the context of the numbers around it.

The number we want to reach is $$5072$$.

Look at the place–value of each digit:

• Thousands digit = $$5$$
• Hundreds digit = $$0$$
• Tens digit = $$7$$
• Ones digit = $$2$$

Convert every digit into the value it actually contributes to the number:

Value from thousands place: $$5 \times 1000 = 5000$$

Value from hundreds place: $$0 \times 100 = 0$$

Value from tens place: $$7 \times 10 = 70$$

Value from ones place: $$2 \times 1 = 2$$

Add these four contributions:

\[ (5 \times 1000) + (0 \times 100) + (7 \times 10) + (2 \times 1) = 5072 \]

Thus a third, different way to obtain $$5072$$ is

$$(5 \times 1000) + (7 \times 10) + (2 \times 1) = 5072$$

Step 1 – Recall the value of one lakh
A lakh is written as $$1\,00\,000$$, which is the same as $$10^5$$.

Step 2 – Write one thousand in powers of 10
One thousand is $$1\,000 = 10^3$$.

Step 3 – Find the value of a thousand lakh
A thousand lakh means $$1\,000 \times 1\text{ lakh}$$.
Substituting the numerical values:
$$1\,000 \times 1\text{ lakh} = 10^3 \times 10^5.$$

Step 4 – Use the law of indices
When we multiply two powers of 10, we add the exponents: $$10^3 \times 10^5 = 10^{3+5} = 10^8.$$

Step 5 – Count the zeros
The number $$10^8$$ is the digit 1 followed by 8 zeros: $$100\,000\,000$$.

Conclusion
Therefore, a thousand lakh contains eight zeros.

In words, “a hundred thousand” means one hundred groups of one thousand.

First write one thousand in numerals:

$$1\,000$$

Multiplying by one hundred gives the value of a hundred thousand:

Thus a hundred thousand is written as $$100\,000$$.

Examine its digits:

• The left-most digit is $$1$$ (hundred-thousands place).
• The remaining digits are zeros.

Counting those zeros:

$$0,0,0,0,0 \;\Rightarrow\; 5 \text{ zeros}$$

Therefore, a hundred thousand has five zeros.

Step 1 · Locate the acceptable interval for each kind of rounding.

Step 2 · Find the common part of all five intervals.
The tightest (smallest) interval is the one for rounding to the nearest ten: from $$4,99,99,995$$ up to $$5,00,00,004$$.
Because every other interval is wider, every number in this ten-number strip automatically belongs to all the other intervals as well.

Step 3 · State the required numbers.
Hence the possible numbers are

$$4,99,99,995,\;4,99,99,996,\;4,99,99,997,\;4,99,99,998,\;4,99,99,999,$$
$$5,00,00,000,\;5,00,00,001,\;5,00,00,002,\;5,00,00,003,\;5,00,00,004.$$

Step 4 · Count them.
There are 10 consecutive whole numbers in the list.

First recall two facts you already know about whole numbers:

• Multiplication is commutative: $$a \times b = b \times a$$.
• When we write a fraction such as $$\tfrac{10}{2}$$, the horizontal bar means “divide 10 by 2”. Thus $$\tfrac{10}{2}=5$$ because 10 contains two 5’s.

Using these ideas we can re-write the number 5:

This is simply another way of saying “half of 10”.

What happens if we multiply an arbitrary number $$N$$ by 5?

Replace 5 by $$\tfrac{10}{2}$$ and use the commutative and associative laws (they allow us to change the order of multiplying and dividing without changing the result):

$$N \times 5 = N \times \frac{10}{2}$$

$$\phantom{N \times 5}= \frac{N}{2} \times 10$$    (first divide by 2, then multiply by 10)

Thus multiplying by 5 gives exactly the same answer as “divide by 2 and then multiply by 10”.

Checking with the two examples

Example 1 (Roxie)

$$116 \times 5 = \frac{116}{2} \times 10 = 58 \times 10 = 580$$ — the same working Roxie used.

Example 2 (Estu)

Before repeating the steps, notice that Estu used the very similar fact

so

$$824 \times 25 = \frac{824}{4} \times 100 = 206 \times 100 = 20600,$$

again “divide first, then attach two zeros”.

Why does the trick feel easier?

• Dividing by 2 is just “halving”, which is quick to do mentally.
• Multiplying by 10 simply appends one zero at the end of a whole number.

Because 5 is half of 10, the two-step process uses only the simplest mental operations, yet by the rules of arithmetic it always equals the original single multiplication by 5.

Question 1 – Patterns in products

We evaluate the given products first, look carefully at “what is happening’’ in the digits and then continue the pattern in the same way.

(i)  Products of 1-digits made only of 1

Step 1 : evaluate the three products.

• $$11\times11=121$$
• $$111\times111=12321$$
• $$1111\times1111=1234321$$

Step 2 : observe – the digits first climb up from $$1$$ to the number of 1’s and then climb down symmetrically. In other words

Step 3 : extend it – for four and five 1’s.

• $$11111\times11111=123454321$$
• $$111111\times111111=12345654321$$

(ii)  Products beginning with 66 …

Step 1 : evaluate the three products.

• $$66\times61=4026$$
• $$666\times661=440226$$
• $$6666\times6661=44402226$$

Step 2 : observe.

• Count of $$4$$’s in the answer = one less than the count of $$6$$’s in the first number.
• There is a single $$0$$ in the middle.
• Count of $$2$$’s = count of $$4$$’s, and the last digit is always $$6$$.

Step 3 : extend for five 6’s.

• $$66666\times66661=4444022226$$

(iii)  Products of only 3’s with the same digits ending in 5

Step 1 : evaluate.

• $$3\times5=15$$
• $$33\times35=1155$$
• $$333\times335=111555$$

Step 2 : observe – in every product the first half is made of $$1$$’s, the second half of $$5$$’s, and both halves have the same length.

Step 3 : extend for four 3’s.

• $$3333\times3335=11115555$$

(iv)  Squares just above 100

Step 1 : evaluate.

• $$101\times101=10201$$
• $$102\times102=10404$$
• $$103\times103=10609$$

Step 2 : observe – let the number be $$100+k$$.

• The first three digits are $$100+2k$$.
• The last two digits are $$k^2$$ (written with two places).

Step 3 : extend to $$104$$ and $$105$$.

• $$104\times104=(100+4)^2=10816$$
• $$105\times105=(100+5)^2=11025$$

Thus, recognising the patterns lets us write the next answers almost instantly.

Step 1 – What “number of digits” means. A positive whole number has as many digits as the places it occupies in the place-value chart. For example, $$537$$ has $$3$$ digits and $$10{,}472$$ has $$5$$ digits.

Step 2 – Name the digit counts. Suppose the first number has $$m$$ digits and the second number has $$n$$ digits.

Step 3 – Smallest and largest numbers with a given number of digits.

• The smallest $$m$$-digit number is $$10^{m-1}$$ (for $$m=3$$ this is $$10^{2}=100$$).
• The largest $$m$$-digit number is $$10^{m}-1$$ (for $$m=3$$ this is $$10^{3}-1=999$$).
• Similarly, the smallest $$n$$-digit number is $$10^{n-1}$$ and the largest is $$10^{n}-1$$.

Step 4 – Smallest possible product. The smallest product comes from multiplying the two smallest factors:

The number $$10^{m+n-2}$$ is a $$1$$ followed by $$(m+n-2)$$ zeros, so it has exactly $$(m+n-1)$$ digits. Hence no product can have fewer than $$(m+n-1)$$ digits.

Step 5 – Largest possible product. The largest product comes from multiplying the two largest factors, $$10^{m}-1$$ and $$10^{n}-1$$. Expanding the brackets,

Since $$m \ge 1$$ and $$n \ge 1$$, we have $$10^{m} \ge 10$$ and $$10^{n} \ge 10$$, so

We are therefore subtracting a positive quantity from $$10^{m+n}$$, which gives the upper bound

The number $$10^{m+n}$$ is a $$1$$ followed by $$(m+n)$$ zeros — the smallest $$(m+n+1)$$-digit number. Any product stays strictly below it, so the product has at most $$(m+n)$$ digits.

Step 6 – Combine the two bounds. Putting Steps 4 and 5 together, every product $$P$$ of an $$m$$-digit number and an $$n$$-digit number satisfies

Whole numbers from $$10^{m+n-2}$$ up to $$10^{m+n-1}-1$$ have exactly $$(m+n-1)$$ digits, and whole numbers from $$10^{m+n-1}$$ up to $$10^{m+n}-1$$ have exactly $$(m+n)$$ digits. So the product must have either $$(m+n-1)$$ or $$(m+n)$$ digits — no other count is possible.

Step 7 – Check with examples.

Step 8 – Conclusion. There is a definite connection: if one factor has $$m$$ digits and the other has $$n$$ digits, their product has either $$(m+n-1)$$ digits or $$(m+n)$$ digits — never any other count.

Step 1 – Identify the range of every 2-digit number
Any 2-digit whole number is at least $$10$$ and at most $$99$$.

Step 2 – Name the two numbers and write an inequality
Let the two numbers be $$a$$ and $$b$$.
Then $$10 \le a \le 99$$ and $$10 \le b \le 99.$$

Step 3 – Find the smallest possible product
The smallest product occurs when both numbers are the smallest 2-digit number:
$$a = 10,\; b = 10 \;\Rightarrow\; ab = 10 \times 10 = 100.$$
Since $$100$$ has three digits, the product can never be smaller than a 3-digit number.

Step 4 – Find the largest possible product
The largest product occurs when both numbers are the largest 2-digit number:
$$a = 99,\; b = 99 \;\Rightarrow\; ab = 99 \times 99 = 9801.$$
The number $$9801$$ has four digits and is still less than $$10000$$, the smallest 5-digit number. Hence a 5-digit product is impossible.

Step 5 – Write the complete inequality for every possible product
Because multiplication preserves the order for positive numbers,
$$100 \le ab \le 9801.$$(Every possible product lies between these two inclusive limits.)

Step 6 – Translate the inequality into number of digits
• All numbers from $$100$$ to $$999$$ are 3-digit numbers.
• All numbers from $$1000$$ to $$9801$$ are 4-digit numbers.
Therefore every possible product is either a 3-digit or a 4-digit number—never 2-digit and never 5-digit.

Step 7 – Check with a few concrete examples

All sample products agree with the conclusion.

Conclusion
Since every possible product lies between $$100$$ and $$9801$$, it must contain exactly three or four digits. Roxie’s statement is correct.

Step 1 – Understand the range of numbers involved
Every 2-digit whole number lies between $$10$$ and $$99$$ (both inclusive).

Step 2 – Locate the largest possible product
If we want the greatest product that two such numbers can give, we multiply the two greatest 2-digit numbers: $$99 \times 99$$.

Step 3 – Find that product once
Carrying out the multiplication very quickly (long multiplication may be shown on the board if required):

$$99 \times 99 = 9801$$

Step 4 – Use the result to test Roxie’s claim
• $$9801$$ is the largest product that can ever occur.
• If Roxie’s statement is “the product of any two 2-digit numbers is less than $$10\,000$$ (or has at most four digits, etc.)”, we have just shown the worst-case product and it still satisfies the condition.
• Every other pair of 2-digit numbers will give a product $$\le 9801$$, so they will automatically satisfy the same condition.

Step 5 – Answer the question asked
Therefore, we do not have to try all $$90 \times 90 = 8100$$ different pairs. Checking the single, logically chosen ‘extreme’ case $$99 \times 99$$ is enough to confirm (or reject) Roxie’s claim.

Let the two three-digit numbers be $$a$$ and $$b$$.

Because each is a three-digit number,

$$100 \le a \le 999 \quad\text{and}\quad 100 \le b \le 999.$$

Smallest possible product

When both numbers take their smallest value, $$a=100$$ and $$b=100$$, so

$$a \times b = 100 \times 100 = 10\,000.$$

The number $$10\,000$$ has 5 digits, not 4.

Largest possible product

When both numbers take their greatest value, $$a=999$$ and $$b=999$$, so

$$a \times b = 999 \times 999 = 998\,001,$$

which has 6 digits.

Therefore any product $$a\times b$$ must satisfy

$$10\,000 \le a\times b \le 998\,001.$$

Every number in this range has either 5 or 6 digits. A 4-digit result (from $$1\,000$$ to $$9\,999$$) never occurs.

Hence multiplying one three-digit number by another can never give a four-digit number.

We want to know whether the product of a 4-digit number and a 2-digit number can have exactly 5 digits.

First note what each term means:

• A 4-digit number lies between $$1000$$ and $$9999$$ (both inclusive).
• A 2-digit number lies between $$10$$ and $$99$$ (both inclusive).

Therefore the smallest possible product is obtained by multiplying the two smallest numbers in each range:

\[1000 \times 10 = 10000\]

$$10000$$ has exactly 5 digits.

The largest possible product is obtained by multiplying the two largest numbers in each range:

\[9999 \times 99 = 9999 \times (100 - 1) = 9999 \times 100 - 9999 = 999900 - 9999 = 989901\]

$$989901$$ has 6 digits.

Thus every product lies between $$10000$$ and $$989901$$. This means:

• The product will always have at least 5 digits (never 4 or fewer);
• Sometimes it will have 5 digits (for example $$1000 \times 10 = 10000$$ or $$1234 \times 12 = 14808$$);
• Sometimes it will have 6 digits (for example $$9999 \times 99 = 989901$$).

Hence it is certainly possible to get a 5-digit number when a 4-digit number is multiplied by a 2-digit number.

Let us first decode what “a k-digit number” means.

• The smallest k-digit number is $$10^{k-1}$$ (for example, for k = 3 we get 100).
• The greatest k-digit number is $$10^{k}-1$$ (for example, for k = 3 we get 999).

Suppose we multiply an m-digit number by an n-digit number.

1. Smallest possible product:
     $$10^{m-1}\times10^{n-1}=10^{m+n-2}.$$
     The number $$10^{m+n-2}$$ begins with a 1 followed by m+n−2 zeros, so it has m + n − 1 digits.
2. Largest possible product:
     $$(10^{m}-1)\,(10^{n}-1) \;<\; 10^{m}\times10^{n} \;=\; 10^{m+n}.$$
     Any number that is smaller than $$10^{m+n}$$ but at least $$10^{m+n-1}$$ has $$m+n$$ digits. If the product happens to be smaller than $$10^{m+n-1}$$ it has only $$m+n-1$$ digits instead.

Therefore a product of an m-digit and an n-digit number can have only two possible digit–counts:

This matches every line already printed in the question:

Now fill in the missing entries exactly the same way.

Hence the observed pattern indeed extends to all other cases.

Step 1: Write the numbers clearly.

We have to multiply $$1250$$ by $$380$$.

Step 2: Split one number to make the calculation easier.

Notice that $$380 = 38 \times 10$$. So

$$1250 \times 380 = 1250 \times (38 \times 10).$$

Step 3: Multiply $$1250$$ by $$38$$.

• Method (a) — Distributive law:
    $$1250 \times 38 = 1250 \times (40 - 2)$$
    $$= 1250 \times 40 \; - \; 1250 \times 2$$
    $$= 50\,000 \; - \; 2\,500$$
    $$= 47\,500.$$
• Method (b) — Long multiplication (the same result):
  

  1250

  ×  38

  ────────

  10 000   ← 1250 × 8

  + 37 500   ← 1250 × 30

  ────────

  47 500

Either way we get $$1250 \times 38 = 47\,500$$.

Step 4: Multiply the result by $$10$$.

$$47\,500 \times 10 = 475\,000.$$ This is the required product.

\[\boxed{475\,000}\]

Step 5: Read the number in the two naming systems.

• Indian System (periods: lakh / thousand / hundreds):
    Write it with commas as $$4,75,000$$.
    Read it as “four lakh seventy-five thousand.”
• International (American) System (periods: thousands / hundreds):
    Write it with commas as $$475,000$$.
    Read it as “four hundred seventy-five thousand.”

Thus, according to legend, Purandaradasa composed 475 000 kirtanas — four lakh seventy-five thousand (Indian) or four hundred seventy-five thousand (American).

Given facts from the lesson

• Year of birth of Purandaradasa  $$= 1484\,\text{AD}$$
• Year of death of Purandaradasa $$= 1564\,\text{AD}$$
• Total number of songs composed   $$= 4,75,000$$ (four lakh seventy-five thousand)
• He began to compose when he was 30 years old.

(i) How many years did he live?

Age at death

$$1564-1484 = 80$$

Purandaradasa lived for $$80\,\text{years}$$.

For how many of those years was he composing?

He started at 30, so the number of years spent in composing is

$$80-30 = 50$$

He therefore composed for $$50\,\text{years}$$.

(ii) Average number of songs composed per year

To find the average, divide the total number of songs by the total number of years of composition.

Number of songs per year

$$\frac{4,75,000}{50}$$

First divide by 10 and then by 5 (because $$50 = 10\times5$$):

• $$4,75,000 \div 10 = 47,500$$
• $$47,500 \div 5 = 9,500$$  because  $$5\times9,500 = 47,500$$

Hence the average number of songs composed each year is

\[9,500\]

Answer to the two questions

• He lived for 80 years and began composing at the age of 30.
• He had to compose about 9,500 songs every year.

Step 1 – Separate the non-zero digits from the zeros
$$2100 = 21 \times 100 \;(\text{two zeros})$$
$$70\,000 = 7 \times 10\,000 \;(\text{four zeros})$$

Step 2 – Multiply the non-zero parts
$$21 \times 7 = 147$$

Step 3 – Attach the total number of zeros
Total zeros  = 2 + 4 = 6.
Appending six zeros to 147 gives
\[147,000,000\]

Step 4 – State the distance
So, $$2100 \times 70\,000 = 147\,000\,000\;\text{km}.$$

Step 5 – Read the number in the two naming systems

Thus the approximate Earth–Sun distance obtained from the product is 147 million kilometres, or 14 crore 70 lakh kilometres.

Step 1 – What quantity do we already know?
Scientists had determined the speed of light. A very accurate modern value is
$$v = 3\times10^5\;\text{km s}^{-1}$$ (that is, light covers three hundred thousand kilometres every single second).

Step 2 – What extra observation is needed?
The next job is to find how long a ray of sunlight takes to reach the Earth. Precise astronomical measurements show that the sunlight we see now actually left the Sun a little more than eight minutes ago. The commonly accepted value is “8 minutes 20 seconds”.
Changing this mixed time into seconds only (because the speed of light is given per second):
\[8\times60 = 480\;\text{s}\]
Adding the extra 20 s gives
$$t = 480 + 20 = 500\;\text{s}.$$

Step 3 – Use the basic distance formula
Distance, speed and time are related by
$$\text{distance} = \text{speed}\times\text{time}$$ So for the Earth–Sun separation
$$d = v\,t = \bigl(3\times10^5\;\text{km s}^{-1}\bigr)\,(500\;\text{s}).$$

Step 4 – Do the arithmetic carefully
First multiply the ordinary numbers: $$3\times500 = 1500$$.
Next deal with the powers of ten: $$10^5\times10^0 = 10^5$$ (because 500 has no power of ten factor written).
Therefore
$$d = 1500\times10^5\;\text{km}.$$

Step 5 – Write the answer in scientific notation
“1500” is not between 1 and 10, so take out another factor 10³: $$1500 = 1.5\times10^3.$$
Hence $$d = \bigl(1.5\times10^3\bigr)\times10^5 = 1.5\times10^{3+5}\;\text{km} = 1.5\times10^8\;\text{km}.$$

Conclusion
Combining a measured time lag of about 500 s with the known speed of light gives the distance between the Earth and the Sun as approximately

This method—timing how long sunlight (or radio signals reflected from another planet) takes to arrive—was the first to give a value close to today’s accepted figure, and it is still the basic idea behind the modern laser-radar techniques used for interplanetary distance measurements.

The problem asks us to evaluate the product $$6400 \times 62\,500$$.

Step 1 : Split a factor to simplify the calculation.

Notice that $$6400 = 64 \times 100$$. Therefore

$$6400 \times 62\,500 = (64 \times 100) \times 62\,500 = 64 \times 62\,500 \times 100.$$

Step 2 : Find $$64 \times 62\,500$$.

Write $$64 = 8 \times 8$$:

$$64 \times 62\,500 = 8 \times 8 \times 62\,500.$$

• First multiplication: $$8 \times 62\,500 = 500\,000.$$
• Second multiplication: $$8 \times 500\,000 = 4\,000\,000.$$

So $$64 \times 62\,500 = 4\,000\,000.$$

Step 3 : Multiply by the remaining 100.

$$4\,000\,000 \times 100 = 400\,000\,000.$$

Conclusion.

The required product is

\[400\,000\,000\]

In words, the Amazon River discharges four hundred million litres of water—i.e. 40 crore litres—into the Atlantic Ocean every second.

We have to find the value of the division
$$13,95,000 \div 150$$.

Step 1  Split the divisor.
Since $$150 = 15 \times 10$$, we can divide in two stages:

$$13,95,000 \div 150 = \bigl(13,95,000 \div 10\bigr) \div 15$$

Step 2  First divide by 10.
Dividing by 10 simply removes one zero from the right-hand side:

$$13,95,000 \div 10 = 1,39,500$$

Step 3  Now divide 1,39,500 by 15.

We use long division:

Explanation of the steps:

• 139 divided by 15 gives 9; 9 × 15 = 135. Subtract to get a remainder 4.
• Bring down the next digit 5 to make 45. 45 ÷ 15 = 3; 3 × 15 = 45. Remainder 0.
• The last two zeros come straight down, so we write them in the quotient.

Thus $$1,39,500 \div 15 = 9,300$$.

Step 4  Write the final quotient.

\[ 13,95,000 \div 150 = 9,300 \]

Therefore the distance of the longest single-train journey in the world is 9,300 kilometres.

We have to find the value of the quotient in the statement $$10,50,00,000 \div 700$$.

Step 1 : Rewrite the dividend in the International system.
In the Indian system, the commas in 10,50,00,000 represent “crore–lakh–thousand”.
So

$$10,50,00,000 = 105,000,000.$$

Step 2 : Break the divisor.
Write 700 as a product that is easy to divide by:

$$700 = 7 \times 100.$$

Step 3 : Divide by 100 first.

$$105,000,000 \div 100 = 1,050,000.$$

Step 4 : Now divide the result by 7.

$$1,050,000 \div 7 = 150,000,$$ because $$7 \times 150,000 = 1,050,000.$$

Step 5 : Write the final quotient.

\[105,000,000 \div 700 = 150,000\]

Therefore, an adult blue whale can weigh more than 150 000 kilograms.

Problem Restatement
Find the quotient when the dividend $$52,00,00,00,000$$ is divided by the divisor $$130$$.

Step 1 – Remove the Indian commas
$$52,00,00,00,000 = 52\,000\,000\,000$$ (fifty-two billion).

Step 2 – Cancel a common factor of 10
Both numbers end in one zero, so divide each by 10:

\(52\,000\,000\,000 \div 130\;=\;(52\,000\,000\,000 \div 10) \div (130 \div 10)\)

\(=\;5\,200\,000\,000 \div 13\).

Step 3 – Long division of \(5\,200\,000\,000\) by \(13\)

1. 13 goes into 52 exactly 4 times (because $$13\times4 = 52$$). Write 4 as the first digit of the quotient.
2. The remainder is 0, and eight zeros are still to be brought down. Each of these zeros gives 0 in the quotient.

Thus the quotient is $$4\,0\,0\,0\,0\,0\,0\,0\,0 = 400\,000\,000$$.

We may show the entire calculation in one line:

Step 4 – Write the answer in both systems

• International: 400 million  (\(400\,000\,000\)).
• Indian: 40 crore  (\(40,00,00,000\)).

Verification
Multiply back to check: $$130 \times 400\,000\,000 = 52\,000\,000\,000,$$ confirming the result.

Quotient: $$40,00,00,000$$ (that is, $$400\,000\,000$$) tonnes.

Facts given in the question

• Number of passengers one RMS Titanic can carry = $$2500$$
• Number of such ships considered = $$5000$$
• Population of Mumbai (approx.) = $$1,25,00,000$$

Step 1  Find the total carrying capacity of 5000 ships

$$\text{Total capacity}=2500 \times 5000$$

Write each number as a product of a small number and powers of 10:

$$2500 = 25 \times 100$$
$$5000 = 5 \times 1000$$

Multiply:

$$2500 \times 5000 = (25 \times 100)(5 \times 1000) = 25 \times 5 \times 100 \times 1000$$

First the non-zero digits:

$$25 \times 5 = 125$$

Now attach all the zeros (two from 100 and three from 1000):

$$12500000$$

Thus,

passengers can be carried by 5000 ships.

Step 2  Compare with Mumbai’s population

Total capacity of 5000 ships = $$1,25,00,000$$ passengers

Population of Mumbai = $$1,25,00,000$$ people

Conclusion

Since both numbers are equal, the whole population of Mumbai can be accommodated in 5000 RMS Titanic ships.

Previous exercise value for the Earth–Sun distance:

$$d = 150\,000\,000\text{ km}$$

Speed of travel given in the question:

$$v = 1\,000\text{ km per day}$$

1. Find the total number of days needed.

$$\text{Number of days} = \dfrac{d}{v} = \dfrac{150\,000\,000\text{ km}}{1\,000\text{ km/day}} = 150\,000\text{ days}$$

2. Convert these days into years (taking $$1\text{ year} = 365\text{ days}$$).

$$\text{Number of years} = \dfrac{150\,000\text{ days}}{365\text{ days/year}} \approx 410.96\text{ years} \;\approx\; 411\text{ years}$$

3. Compare with a human lifetime.

A generous human lifetime is about $$80\text{--}100$$ years, which is far less than $$411$$ years.

Conclusion: Travelling $$1\,000\text{ km}$$ every day, you would need about $$411$$ years to reach the Sun, so you cannot do it within a single human lifetime.

Below is a set of five self-made “large–number” problems together with fully worked solutions that you can share with your classmates.

Fun Question 1 — Population Difference

According to a recent census, the population of China is $$1\,439\,323\,776$$ and that of India is $$1\,352\,642\,280$$. How many more people live in China than in India? Also write that difference in words.

Step 1 Arrange the two numbers one under the other and subtract.

$$\begin{array}{r} 1\,439\,323\,776\\[-2pt] -\;1\,352\,642\,280\\ \hline \phantom{0}\,086\,681\,496 \end{array}$$

Step 2 State the answer.

The difference is $$86\,681\,496$$.

Step 3 Write in words.

$$86\,681\,496$$ is read as Eighty-six million six hundred eighty-one thousand four hundred ninety-six.

Fun Question 2 — A Beam of Light from the Sun

The average distance from the Sun to the Earth is about $$149\,600\,000\ \text{km}$$. Light travels at about $$300\,000\ \text{km s}^{-1}$$. How long (in minutes and seconds) does a ray of sunlight take to reach us?

Step 1 Write the formula.

$$\text{Time} = \dfrac{\text{Distance}}{\text{Speed}}$$

Step 2 Substitute the values.

$$\text{Time} = \dfrac{149\,600\,000}{300\,000}\;\text{s}$$

Step 3 Simplify the fraction.

Divide numerator and denominator by $$1\,000$$:

$$\dfrac{149\,600}{300} = \dfrac{1\,496}{3} = 498.666\dots\ \text{s}$$

Step 4 Convert seconds to minutes and seconds.

$$498\,\text{s} = 8\times60\,\text{s} + 18\,\text{s}$$ because $$8\times60 = 480$$ and $$498-480 = 18$$.

So the light takes 8 minutes 18 seconds (≈8 min 19 s when rounded to the nearest second).

Fun Question 3 — School Recycling Drive

A school recycles $$250\,000\ \text{g}$$ of paper every day.

1. Convert this daily amount into kilograms.
2. How many kilograms of paper are recycled in one week?

Solution (i)

Since $$1\ \text{kg} = 1\,000\ \text{g}$$,

$$250\,000\ \text{g} = \dfrac{250\,000}{1\,000}\ \text{kg} = 250\ \text{kg}$$.

Solution (ii)

There are $$7$$ days in a week, so

$$\text{Weekly total} = 250\ \text{kg}\times7 = 1\,750\ \text{kg}$$.

Fun Question 4 — How Tall Is a Stack of Paper?

One sheet of paper is about $$0.1\ \text{mm}$$ thick. If you stack $$750\,000$$ such sheets,

1. What will be the height of the stack in metres?
2. Is this taller than a 20-storey building if one storey is $$3.5\ \text{m}$$ high?

Step 1 Find the total thickness in millimetres.

$$750\,000\times0.1\ \text{mm} = 75\,000\ \text{mm}$$

Step 2 Convert millimetres to metres.

$$75\,000\ \text{mm} = \dfrac{75\,000}{1\,000}\ \text{m} = 75\ \text{m}$$

Step 3 Compare with the building.

Height of 20 storeys $$=20\times3.5\,\text{m}=70\,\text{m}$$.

Because $$75\,\text{m} > 70\,\text{m}$$, the paper stack is 5 m taller than the building.

Fun Question 5 — Winning the Lottery

You win a jackpot of $$9\,87\,65\,432$$ rupees. The amount is to be paid out in equal monthly instalments over $$7$$ years.

1. How many monthly instalments are there in all?
2. How many rupees will you receive each month? Is any money left over?

Solution (i)

$$7\,\text{years} = 7\times12 = 84\,\text{months}$$.

Solution (ii) Divide the total prize by $$84$$.

$$\dfrac{98\,765\,432}{84} = 1\,175\,778\;\text{remainder}\;80$$

(You may show the long division on the board for your class.)

So you receive ₹ 1,175,778 every month, and at the end there will be ₹ 80 left over which can be added to the final payment.

You can now challenge your friends with these problems or invent similar ones using other large real-life numbers!

Step 1 : Understand the numbers involved.

“One lakh” in the Indian place-value system is written as $$1\,00\,000$$.

Step 2 : Set up the required difference.

Required difference = $$1\,00\,000 - 75\,000$$.

Step 3 : Subtract.

Think of each number as a multiple of one thousand:

$$1\,00\,000 = 100 \times 1000, \qquad 75\,000 = 75 \times 1000.$$

So

$$1\,00\,000 - 75\,000 = (100 \times 1000) - (75 \times 1000) = (100 - 75) \times 1000 = 25 \times 1000.$$

Compute the product:

Conclusion: The population of 75,000 is $$25\,000$$ less than one lakh.

Recall that one lakh means one hundred thousand, that is
$$1\text{ lakh}=1\times100\,000=100\,000.$$

The estimated population is
$$\text{population}=1\,06\,000.$$

To find how much more this number is than one lakh, subtract one lakh from the given population:

$$\text{excess}=\text{population}-1\text{ lakh}=106\,000-100\,000.$$

Do the subtraction digit-wise:

• At the thousands place: $$106-100=6$$ thousands, i.e. $$6\,000.$$
• No other places change because the digits in the ten-thousand and higher places cancel out.

So,

Hence, the population of Chintamani is $$6,000$$ more than one lakh.

Given (from the previous two parts of the exercise)

• Population of Chintamani in 2011 = $$75,000$$
• Estimated population of Chintamani in 2024 = $$1,06,000$$

What we have to find

The increase in population from 2011 to 2024, that is

$$\text{Increase} = \text{Population in 2024} - \text{Population in 2011}.$$

Step 1 — Set up the subtraction.

$$1,06,000 - 75,000.$$

Step 2 — Subtract column by column from the right.

Working from the right:

• Ones, tens, hundreds: $$0 - 0 = 0$$ in each column.
• Thousands: $$6 - 5 = 1$$.
• Ten-thousands: $$0 - 7$$ is not possible. Borrow $$1$$ lakh ($$=10$$ ten-thousands). Now $$10 - 7 = 3$$ ten-thousands.
• Lakhs: after the borrow, $$1 - 1 = 0$$.

Step 3 — Read the answer.

$$1,06,000 - 75,000 = 31,000.$$

Therefore, the population of Chintamani increased by $$31,000$$ between 2011 and 2024.

Understanding the situation

A 3-digit number can be written in the form $$N = 100h + 10t + u$$ where

• $$h,t,u$$ are the digits in the hundreds, tens and ones places respectively,
• $$1 \le h \le 9$$ (because the hundreds digit of a 3-digit number cannot be $$0$$) and $$0 \le t,u \le 9$$.

Imagine we start from the display $$000$$ and the only thing a single button press does is to increase one chosen digit by $$1$$.
That means:

• One press raises the chosen digit by exactly $$1$$.
• After nine presses on the same digit it reaches $$9$$ and can go no higher.
• Altogether we must make exactly 30 presses.

Let

$$h+t+u = P$$

denote the total number of useful (i.e. digit-raising) presses. Because a digit cannot exceed $$9$$ we have the restriction

$$0 \le h,t,u \le 9\;.$$

We are free to waste presses by doing a move that is immediately cancelled (for example press “up” and then “down”); therefore we only have to make sure that

$$P \le 30.$$

The job now is to choose $$h,t,u$$ so that

• $$N = 100h + 10t + u$$ is as large as possible, and
• $$N$$ is as small as possible,

while respecting $$0 \le h,t,u \le 9$$ and $$h \ge 1$$.

(A) Largest 3-digit number

Because each press placed on the hundreds digit raises the overall number by $$100$$, on the tens digit by $$10$$ and on the ones digit by only $$1$$, we obviously want to spend our useful presses in the following order:

1. Fill the hundreds digit up to $$9$$ (needs $$9$$ presses).
2. Fill the tens digit up to $$9$$ (another $$9$$ presses; running total $$18$$).
3. Fill the ones digit up to $$9$$ (another $$9$$ presses; running total $$27$$).

So with $$27$$ useful presses we have already reached

$$N = 999.$$

There are still $$30-27 = 3$$ presses left. These can be harmlessly wasted (for example, press a digit up once and immediately press it down once, repeating as many times as required), leaving the display unchanged. Hence the largest attainable 3-digit number is

(B) Smallest 3-digit number

To make the number as small as possible we do exactly the opposite: press the hundreds place only once and leave the other two digits at $$0$$. That costs just

$$h = 1,\; t = 0,\; u = 0 \;\Rightarrow\; P = 1.$$

The remaining $$29$$ presses can again be wasted in cancelling pairs, so the final display stays

Since any 3-digit number must be at least $$100$$, this is indeed the smallest possible.

Conclusion

The requirements can be met, and we obtain:

• Largest 3-digit number: $$999$$
• Smallest 3-digit number: $$100$$

Let the three drums of the place-value machine be :

• H-drum  →  every click puts 1 hundred, i.e. $$100$$.
• T-drum  →  every click puts 1 ten, i.e. $$10$$.
• U-drum  →  every click puts 1 one, i.e. $$1$$.

If we finally wish to read the number $$997$$ we must decide how many clicks are given to each drum.

Suppose the required numbers of clicks are $$h$$, $$t$$ and $$u$$ respectively. After all the clicks the displayed number will be

$$100h+10t+u.$$

Therefore the drums have to satisfy

$$100h+10t+u = 997\[4pt] \Rightarrow 10h+t = 99\;\text{ and }\;u=7.$$

The digits shown on any drum can only be $$0,1,2,\ldots ,9$$, so

$$0\le h\le 9,\;0\le t\le 9,\;u=7.$$

The equation $$10h+t=99$$ must now be solved under this restriction. Trying all possible values of $$h$$ from $$0$$ to $$9$$ we get

The only admissible solution is

$$h = 9,\; t = 9,\; u = 7.$$

Thus the number of clicks must be

$$h+t+u = 9+9+7 = 25.$$

No other triple of digits satisfies the place-value condition, so 25 clicks is the one and only way to make 997.

What does “one button click” mean?
Every time we press any key on a calculator – a digit (0–9), an operation symbol (+, –, ×, ÷), a decimal point, the EXP / 10^x key, the = key, … – we make one click. The aim is to reach the desired display with as few such clicks as possible.

In the previous exercise we had four numbers

• (i) $$87595762$$
• (ii) $$8547280$$
• (iii) $$99900046$$
• (iv) $$100000000$$

We examine each number and decide which key–sequence is shortest.

1. $$87595762$$
   The number has 8 digits and no long run of zeros at the end. Any alternative such as $$8759576.2 \times 10$$ or $$8759 \times 10000 + 5762$$ needs more than 8 clicks (because of extra operation keys, the decimal point, etc.).
   Minimal sequence: type every digit once.
   Expression to show what we really did: $$87595762$$ (no operation symbol is needed).

2. $$8547280$$
   This is a 7-digit number with only one zero at the end. Typing it directly costs 7 clicks. Any indirect method (for example $$854728 \times 10$$ or repeated “×10, =”) takes at least 9 clicks.
   Minimal sequence: 7 successive digit-keys.
   Expression: $$8547280$$.

3. $$99900046$$
   Eight digits again, and the two zeros are in the middle. Direct typing therefore still needs only 8 clicks. Splitting the number – say $$999 \times 100001 - (something)$$ – would certainly exceed that. Hence the straight method wins.
   Minimal sequence: press the eight digit-keys.
   Expression: $$99900046$$.

4. $$100000000$$
   This number is special: it is a pure power of 10, namely $$10^8$$. If our calculator has the EXP (or 10^x) key, just three clicks are enough:

   • 1
   • EXP (or 10^x)
   • 8

   That is far fewer than the nine digit-clicks we would need to type 1 followed by eight zeros.
   Minimal sequence: 3 clicks.
   Expression to be written on the answer sheet: $$10^8$$.

Summary of the shortest expressions

Thus, except for the last (which is a neat power of 10), the quickest way is simply to press every digit once – that already needs the smallest possible number of button clicks.

The display of the counter is initially 00000. Each of the five wheels has its own button. One click on a button advances only that wheel by one step (0→1→2→…→9→0). Therefore, to turn a wheel that is showing 0 into a wheel that shows the digit $$d$$ (where $$d\in\{0,1,\ldots ,9\}$$) we must click its button exactly $$d$$ times — clicking either fewer or more times cannot leave the wheel at the required digit.

Let the required number be written in expanded (place-value) form

$$N = d_4\times10^4 + d_3\times10^3 + d_2\times10^2 + d_1\times10 + d_0,$$

where $$(d_4,d_3,d_2,d_1,d_0)$$ are the digits in the ten-thousands, thousands, hundreds, tens and ones places respectively. To obtain the display $$d_4d_3d_2d_1d_0$$ from 00000 we have to make

• $$d_4$$ clicks on the ten-thousands button,
• $$d_3$$ clicks on the thousands button,
• $$d_2$$ clicks on the hundreds button,
• $$d_1$$ clicks on the tens button, and
• $$d_0$$ clicks on the ones button.

Because every wheel works independently, the total number of clicks is simply the sum of these individual counts:

\[\boxed{\text{Minimum clicks required}=d_4+d_3+d_2+d_1+d_0}.\]

But $$d_4+d_3+d_2+d_1+d_0$$ is nothing but the sum of the digits of $$N$$. Hence,

The smallest number of button clicks needed to obtain any particular number equals the sum of the digits of that number.

This is exactly the connection you can observe in the table: for every listed number, its ‘minimum clicks’ column shows the sum of its digits.

Step 1 ‣ Recall what “least button clicks” meant
On a basic calculator we must press one key for every printed symbol. To avoid pressing the zero key many times we rewrite a large number as its expanded form using powers of 10, for example

\[72\,45\,18\,000=7×10^{7}+2×10^{6}+4×10^{5}+5×10^{4}+1×10^{3}+8×10^{2}\]

This way only the non-zero digits and the symbols “×”, “+” and “10” are pressed; every string of zeros is replaced by one power of 10.

Step 2 ‣ Write the same number in Indian place value notation

Indian groups are crore-lakh-thousand-hundred-ones:

• $$7×10^{7}=7$$ crore
• $$2×10^{6}=2$$ ten-lakh = 20 lakh
• $$4×10^{5}=4$$ lakh
• $$5×10^{4}=5$$ ten-thousand = 50 thousand
• $$1×10^{3}=1$$ thousand
• $$8×10^{2}=8$$ hundreds

Combining these we read $$72\,45\,18\,000$$ exactly as
“Seventy-two crore forty-five lakh eighteen thousand”.

Step 3 ‣ Why do the two writings always coincide?

1. Every digit of a whole number sits in a place worth a power of 10:
   $$\text{digit}×10^{n}$$
2. The “least-click” expression is nothing but the sum of all such $$\text{digit}×10^{n}$$ terms; this is precisely the expanded form taught in the place-value table.
3. The Indian system merely groups the same powers of 10 as
   crore ($10^{7},10^{6}$), lakh ($10^{5},10^{4}$), thousand ($10^{3},10^{2}$) and so on. Hence every term you type on the calculator already names one of those groups.

Conclusion
Because both methods break the number into its digits multiplied by the correct powers of 10, the calculator’s “least button clicks” expression and the Indian place value notation are two ways of writing the very same expanded form. That is why they always match.

Step 1 : Break a 7-digit number into Indian place–value periods
Every 7-digit number is written as
\[\text{(two–digit)}\;\text{lakhs}\;\;\text{(two–digit)}\;\text{thousand}\;\;\text{(three–digit units)}\]
so its name has three independent blocks:

• lakhs block : two-digit word + “lakhs”
• thousands block : two-digit word + “thousand”
• units block : three-digit word (of the form “digit-word hundred two-digit-word”)

To get the longest name we must give each block the longest possible word(s).

Step 2 : The longest two-digit words

The greatest total is 12 letters, obtained from any of

• “seventy seven”
• “seventy eight”
• “seventy three”

We shall use “seventy seven”.

Step 3 : The longest hundreds part
The hundreds part is “digit-word hundred two-digit-word”.
The digit-words with the most letters are three, seven, eight (each 5 letters).
Keeping the two-digit word as “seventy seven”, the units block becomes

“seven hundred seventy seven”

Letters in this block: 5 (seven) + 7 (hundred) + 7 (seventy) + 5 (seven) = 24.

Step 4 : Build the full number
Choose “seventy seven” in every possible position:

Number = $$77,77,777$$

Name = “Seventy seven lakhs seventy seven thousand seven hundred seventy seven”.

Step 5 : Count the letters

• seventy (7)
• seven (5) –> 12
• lakhs (5) –> 17
• seventy (7) –> 24
• seven (5) –> 29
• thousand (8) –> 37
• seven (5) –> 42
• hundred (7) –> 49
• seventy (7) –> 56
• seven (5) –> 61

Total letters = 61.

No other choice can beat this because

• 12 is the maximum for any two-digit block, and both lakhs & thousands blocks already have 12.
• 24 is the maximum for a three-digit block (5+7+12), which we have used.

Hence the 7-digit number with the maximum letters in its name is

Step 1  — Understand the requirement

Whenever any two digits of our 9-digit number are interchanged, the new number must be greater than the original.
So the original number must be the smallest possible arrangement that can be formed with its own digits. If it were not the smallest, then swapping the leftmost “too-large” digit with a smaller one appearing later would give a still smaller number, contradicting the condition.

Step 2  — What is the smallest arrangement of given digits?

• The smallest arrangement is obtained by writing the digits in ascending order from left to right.
• If any digit were repeated, exchanging the two identical digits would leave the number unchanged, not larger. Hence all nine digits must be different.
• The left-most digit of a 9-digit number cannot be $$0$$ (otherwise it would not really be 9-digit). Therefore $$0$$ cannot appear at all, because in an ascending list it would necessarily come first.
• The only set of nine distinct, non-zero digits available is $$\{1,2,3,4,5,6,7,8,9\}$$.

Step 3  — Form the number

Writing these digits in increasing order gives

Check: If we swap any pair of digits, a larger digit moves to an earlier position (or a smaller digit moves to a later one), so the new number is certainly bigger. Thus the condition is satisfied.

Step 4  — Count how many such numbers exist

Because every acceptable number must

1. use each of the nine non-zero digits exactly once, and
2. place them strictly in ascending order,

there is only one possible number.

Conclusion

The required 9-digit number is $$123456789$$, and exactly one such number exists.

Step 1 : Understand the task
We have the 20-digit number
$$12345123451234512345$$
and we must strike out exactly 10 digits (keeping the order of the remaining ones) so that the 10 digits left form the largest possible number.

Step 2 :� Why a direct trial does not work
Twenty choose ten is a very big number of possibilities, so we need a method instead of listing every case.

Step 3 :� A sure way to get the greatest number
The greedy or stack rule is:

• Read the digits from left to right.
• Whenever the current digit is bigger than the last digit we kept, strike (delete) that smaller digit if we still have deletions left.
• Stop deleting as soon as 10 digits have been struck out. Copy all the rest exactly as they appear.

This rule never lets a smaller digit stay in front of a larger one when we could still delete it, so the final string is the greatest possible.

Step 4 :� Put the rule to work

Exactly 10 digits have been deleted, and the stack now shows the 10 digits kept.

Step 5 :� Read the result
The remaining digits in order are
$$5\,5\,3\,4\,5\,1\,2\,3\,4\,5$$
Therefore the greatest 10-digit number obtainable is
\[5534512345\]

Check
Count of digits kept = 10, count deleted = 10, and any attempt to make the first, second, … digit larger would violate the deletion limit, so the answer is final.

Let us write the English spellings very carefully and examine every possible kind of “next number”.

Throughout the solution we follow the spelling rules used in the NCERT books, i.e.

• Twenty one, twenty two, … (a hyphen in the book, which we may ignore while picking letters)
• One hundred one, one hundred two, … (the textbook does not insert the word “and”).

For every natural number we look at the set of the different letters that appear in its spelling.
Two numbers will “share” a letter as soon as these two sets overlap.

Step 1   Numbers below 20

All the spellings from zero to nineteen are listed once in the textbook. A direct check gives

Consecutive pairs 0–1, 1–2, … , 18–19 all share at least one of the letters e, o, n, r, s, t.
Hence no required pair occurs below twenty.

Step 2   Numbers from 20 to 99

Every spelling in this range is obtained by sticking one of the eight “tens–words”

$$\{\text{twenty},\;\text{thirty},\;\text{forty},\;\text{fifty},\;\text{sixty},\;\text{seventy},\;\text{eighty},\;\text{ninety}\}$$ with (possibly) one of the nine “ones–words”.

(a) If the two consecutive numbers stay inside the same group of ten – for instance 34 and 35 (both begin with thirty) – the common tens–word already supplies many common letters (34 and 35 share at least the letters t and y). Hence no new pair can appear here.

(b) If the pair jumps from one group of tens to the next – for instance 39 to 40 – we must compare the two tens–words. A glance at the eight spellings shows that any two of them share at least one letter out of the set
$$\{t, y, f, i, r, o\}.$$ (E.g. thirty ∩ forty = {t, r, y}.) So even at a “tens-boundary” the two numbers still share some letter.
Conclusion: no such pair below 100.

Step 3   Numbers from 100 to 999

Each of these begins with the word hundred. Therefore two neighbours that lie in the same century automatically share every letter of hundred.

The only remaining possibility is the jump …99 → …00 (for example 199 → 200).
Yet here the spellings

• 199 = one hundred ninety-nine
• 200 = two hundred

still overlap, because they both contain the letters h, n, d, e and o.

Exactly the same overlap happens at every change of hundred. Hence no pair appears below 1000.

Step 4   Thousands, millions, … (general argument)

From 1000 onwards each natural number is built from the fixed “scale words”

thousand, million, billion, trillion, …

Every one of these scale words contains the letter n. In fact

thousand,   million,   billion,   trillion, …

Consider any two consecutive numbers.

• If they lie in the same «thousand–block», «million–block», … they obviously share the complete scale word that stands in front, hence they share at least the letter n.
• If they lie on the boundary – e.g. 999 999 → 1 000 000 – then the first spelling ends with the scale word (… thousand, … million, …) while the second spelling starts with another spelling that certainly contains the letter n (because one, nine, seven, … all contain n or e, etc.). In every case the two sets overlap in at least the letter n (quite often many more).

Therefore it is impossible to find two successive natural numbers whose English spellings have disjoint letter sets.

Final conclusion

The search will never end – such a pair does not exist. No matter how far you count, any two consecutive numbers will always share at least one common English letter.

We want to express one billion in units of lakhs.

Step 1 — Write each number with powers of 10.

A billion is $$1,000,000,000 = 10^{9}$$.

A lakh is $$1,00,000 = 10^{5}$$.

Step 2 — Divide to find the number of lakhs in one billion.

Number of lakhs in a billion = $$\dfrac{10^{9}}{10^{5}}$$.

Use the law of indices $$\dfrac{10^{m}}{10^{n}} = 10^{m-n}$$:

$$\dfrac{10^{9}}{10^{5}} = 10^{9-5} = 10^{4}.$$

Step 3 — Convert $$10^{4}$$ back to an ordinary number.

$$10^{4} = 10,000.$$

Thus,

One billion equals 10,000 lakhs.

The Statue of Unity is 182 m tall.

1. Convert metres to millimetres

Since $$1\text{ m}=1000\text{ mm}$$, the height in millimetres is

$$182\text{ m}=182\times1000\text{ mm}=182\,000\text{ mm}$$

2. Relate coin thickness to required height

Each coin is $$1\text{ mm}$$ thick. If $$n$$ coins are stacked, the height is $$n\text{ mm}$$.

To reach the statue’s height we need

\[n=182\,000\]

3. Conclusion

Therefore, 182 000 coins must be stacked to equal the height of the Statue of Unity.

We know from the statement that

• distance of the trip  = $$12,000\text{ km}$$
• a grey-headed albatross can fly between $$900\text{ km}$$ and $$1,000\text{ km}$$ in one day

Step 1 — Find the least possible number of days.

If the bird manages the maximum daily distance, that is $$1,000\text{ km}$$, then

$$\text{Number of days} = \frac{12,000}{1,000} = 12$$

Step 2 — Find the greatest possible number of days.

If it manages only $$900\text{ km}$$ in a day, then

$$\text{Number of days} = \frac{12,000}{900} \approx 13.3$$

This is a little over 13 days; we round it to about 13 days for an estimate.

Step 3 — State the approximate time.

The journey would therefore take somewhere between 12 and 13 days.

We can summarise the result as

\[\boxed{\text{Time required } \approx 12\text{ to }13\text{ days}}\]

Given data

• Total distance flown  $$ = 13,560\ \text{km} $$
• Total time taken     $$ = 11\ \text{days} $$

1. Average distance covered per day

We divide the total distance by the number of days:

$$\text{Distance per day}=\frac{13,560}{11}\;\text{km}$$

Long-division step by step:

• 11 goes into 13 → 1 time, remainder 2 ( write 1 )
• Bring down 5 → 25. 11 goes into 25 → 2 times, remainder 3 ( write 2 )
• Bring down 6 → 36. 11 goes into 36 → 3 times, remainder 3 ( write 3 )
• Bring down 0 → 30. 11 goes into 30 → 2 times, remainder 8 ( write 2 )

Quotient $$ = 1\,232 $$ and remainder $$ = 8 $$. Hence

$$\frac{13,560}{11}=1,232+\frac{8}{11}\;\text{km}\approx1,232.73\;\text{km}$$

Approximate distance per day (to the nearest kilometre):  $$\boxed{1,233\;\text{km/day}}$$

2. Average distance covered per hour

First change the time into hours:

Number of hours $$ = 11\,\text{days}\times24\,\text{h/day}=264\,\text{h} $$

Now divide the total distance by the number of hours:

$$\text{Distance per hour}=\frac{13,560}{264}\;\text{km}$$

To simplify this fraction, note that $$264=24\times11$$.

Step 1: divide numerator and denominator by 24:

$$\frac{13,560}{24}=565, \qquad \frac{264}{24}=11$$

So $$\displaystyle\frac{13,560}{264}=\frac{565}{11}\;\text{km}$$

Step 2: divide 565 by 11:

• 11 goes into 56 → 5 times (55), remainder 1; bring down 5 → 15.
• 11 goes into 15 → 1 time, remainder 4.

Hence $$\displaystyle\frac{565}{11}=51+\frac{4}{11}\;\text{km}\approx51.36\;\text{km}$$

Approximate distance per hour (to the nearest kilometre):  $$\boxed{51\;\text{km/h}}$$

Conclusion

The bar-tailed godwit flew about $$1,233\;\text{km}$$ every day, which is roughly $$51\;\text{km}$$ every hour without stopping!

Given data

• Height of Somu’s building (mentioned earlier in the textbook)  =  $$15\,\text{m}$$
• Bald eagles fly between $$4500\,\text{m}$$ and $$6000\,\text{m}$$.
• Mount Everest is about $$8850\,\text{m}$$ high.
• Aeroplanes fly between $$10\,000\,\text{m}$$ and $$12\,800\,\text{m}$$.

To know how many times higher each of these heights is as compared to Somu’s building, we divide the given heights by $$15\,\text{m}$$.

1. Bald eagles
   Minimum factor:
   $$\dfrac{4500}{15}=300$$
   Maximum factor:
   $$\dfrac{6000}{15}=400$$
   So, bald eagles fly about 300 to 400 times higher than Somu’s building.
2. Mount Everest
   $$\dfrac{8850}{15}=590$$
   Mount Everest is therefore about 590 times taller than Somu’s building.
3. Aeroplanes
   Lower limit:
   $$\dfrac{10\,000}{15}=666.\overline{6}\approx 667$$
   Upper limit:
   $$\dfrac{12\,800}{15}=853.\overline{3}\approx 853$$
   Hence, aeroplanes fly roughly 667 to 853 times higher than Somu’s building.

Conclusion
Bald eagles, Mount Everest, and aeroplanes are respectively about 300–400 times, 590 times, and 667–853 times higher than Somu’s building.