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NCERT Solutions Ganita Prakash Chapter 5 & 111Section 5.1 Common Multiples and Common Factors — Figure it Out

Book page 110 Updated on2026-09-05

Q1.
Find all multiples of 40 that lie between 310 and 410.
Answer

Write the 40 table and pick the numbers that fall in between.

40, 80, 120, 160, 200, 240, 280, 320, 360, 400, 440, …

Between 310 and 410 we get

320, 360 and 400
A quicker way: 310 ÷ 40 = 7 remainder 30, so the first multiple after 310 is 40 × 8 = 320. Then keep adding 40: 320, 360, 400. The next one, 440, is already past 410.
Q2.
Who am I? a. I am a number less than 40. One of my factors is 7. The sum of my digits is 8. b. I am a number less than 100. Two of my factors are 3 and 5. One of my digits is 1 more than the other.
Answer

a. The number is 35.

“One of my factors is 7” means the number is a multiple of 7. The multiples of 7 below 40 are:

Multiple of 7714212835
Sum of digits71 + 4 = 52 + 1 = 32 + 8 = 103 + 5 = 8 ✔

b. The number is 45.

Having both 3 and 5 as factors means the number is a multiple of 15. Below 100 these are:

Multiple of 15153045607590
Digits1, 53, 04, 5 ✔6, 07, 59, 0
Differ by 1?NoNoYesNoNoNo
4 and 5 → 5 = 4 + 1 ✔    and    45 = 3 × 15 = 5 × 9 ✔
Q3.
A number for which the sum of all its factors is equal to twice the number is called a perfect number. The number 28 is a perfect number. Its factors are 1, 2, 4, 7, 14 and 28. Their sum is 56 which is twice 28. Find a perfect number between 1 and 10.
Answer

The perfect number is 6.

Factors of 6 → 1, 2, 3, 6
1 + 2 + 3 + 6 = 12 = 2 × 6 ✔

Let us check that no other number from 1 to 10 works:

NumberFactorsSum of factorsTwice the numberPerfect?
41, 2, 478No
51, 5610No
61, 2, 3, 61212Yes ✔
81, 2, 4, 81516No
91, 3, 91318No
101, 2, 5, 101820No
Did you know? Perfect numbers are very rare. The first four are 6, 28, 496 and 8128. No one has ever found an odd perfect number, and no one has been able to prove that none exists!
Q4.
Find the common factors of: a. 20 and 28 b. 35 and 50 c. 4, 8 and 12 d. 5, 15 and 25
Answer

List all the factors of each number and pick out the ones that appear everywhere.

PartFactorsCommon factors
a. 20 and 2820 → 1, 2, 4, 5, 10, 20
28 → 1, 2, 4, 7, 14, 28		1, 2, 4
b. 35 and 5035 → 1, 5, 7, 35
50 → 1, 2, 5, 10, 25, 50		1, 5
c. 4, 8 and 124 → 1, 2, 4
8 → 1, 2, 4, 8
12 → 1, 2, 3, 4, 6, 12		1, 2, 4
d. 5, 15 and 255 → 1, 5
15 → 1, 3, 5, 15
25 → 1, 5, 25		1, 5
Always remember: 1 divides every number, so 1 is a common factor of any set of numbers. That is why 1 appears in all four answers.
Q5.
Find any three numbers that are multiples of 25 but not multiples of 50.
Answer

Write the 25 table and cross out the numbers that also appear in the 50 table.

25 table → 25, 50, 75, 100, 125, 150, 175, 200, …

Three such numbers: 25, 75 and 125. (175, 225, 275 … also work.)

Why the alternate ones survive: 50 = 25 × 2, so a multiple of 25 is also a multiple of 50 only when it is an even number of 25s. So take the odd multiples of 25 — 25 × 1, 25 × 3, 25 × 5, … — and none of them will be a multiple of 50.
Q6.
Anshu and his friends play the ‘idli-vada’ game with two numbers, which are both smaller than 10. The first time anybody says ‘idli-vada’ is after the number 50. What could the two numbers be which are assigned ‘idli’ and ‘vada’?
Answer

The first ‘idli-vada’ happens at the first common multiple of the two numbers. So we need two numbers below 10 whose first common multiple is more than 50.

PairFirst common multipleMore than 50?
6 and 742No
5 and 945No
6 and 918No
7 and 856Yes ✔
7 and 963Yes ✔
8 and 972Yes ✔

Answer: the two numbers could be 7 and 8, or 7 and 9, or 8 and 9.

Why only these three: both numbers are below 10, so their product is at most 9 × 8 = 72. To push the first common multiple past 50, the two numbers must be large and must not share a factor — otherwise the first common multiple drops well below the product. 7, 8 and 9 are pairwise co-prime, so their first common multiples are the products 56, 63 and 72.
Note: the answers printed at the back of the book give only 7, 8 and 8, 9 — the pair 7 and 9 works equally well (63 > 50).
Q7.
In the treasure hunting game, Grumpy has kept treasures on 28 and 70. What jump sizes will land on both the numbers?
Answer

Jumpy needs a common factor of 28 and 70.

Factors of 28 → 1, 2, 4, 7, 14, 28
Factors of 70 → 1, 2, 5, 7, 10, 14, 35, 70

The numbers in both lists are

1, 2, 7 and 14
Jump 7 → 7, 14, 21, 28, 35, 42, 49, 56, 63, 70 ✔
Jump 14 → 14, 28, 42, 56, 70 ✔

Answer: jump sizes 1, 2, 7 and 14.

Shortcut: 28 = 2 × 2 × 7 and 70 = 2 × 5 × 7. The primes shared by both are 2 and 7, so the common factors are 1, 2, 7 and 2 × 7 = 14.
Q8.
In the diagram below, Guna has erased all the numbers except the common multiples. Find out what those numbers could be and fill in the missing numbers in the empty regions.
Answer

The overlap shows 24, 48 and 72. These go up by 24 each time, so the first common multiple of the two numbers is 24.

So we need a pair whose first common multiple is 24. A neat choice is 8 and 12:

Multiples of 8Multiples of 128163240566480882448729612366084common multiples
Multiples of 8 on the left, multiples of 12 on the right; the common multiples 24, 48, 72, 96 sit in the middle.
Multiples of 8 only → 8, 16, 32, 40, 56, 64, 80, 88 …
Multiples of 12 only → 12, 36, 60, 84 …
Common multiples → 24, 48, 72, 96 …

Other pairs also work, because they too have 24 as their first common multiple:

  • 3 and 8 — left: 3, 6, 9, 12, 15, 18, 21, 27 …; right: 8, 16, 32, 40 …
  • 6 and 8 — left: 6, 12, 18, 30, 36, 42 …; right: 8, 16, 32, 40 …
  • 4 and 24, 2 and 24, 12 and 24 — here every multiple of 24 sits in the overlap.
How to be sure: the numbers left in the middle are 24, 48, 72 — the multiples of 24. So whichever pair Guna chose, their first common multiple has to be exactly 24, no smaller and no bigger.
Q9.
Find the smallest number that is a multiple of all the numbers from 1 to 10, except for 7.
Answer

We need the smallest number divisible by 1, 2, 3, 4, 5, 6, 8, 9 and 10. Take the biggest power of each prime that is needed.

To cover 8 we need 2 × 2 × 2
To cover 9 we need 3 × 3
To cover 5 (and 10) we need 5
Smallest number = 2 × 2 × 2 × 3 × 3 × 5 = 8 × 9 × 5 = 360

Check every number:

Divide 360 by1234568910
Answer360180120907260454036

Every division is exact, so 360 is the answer.

Q10.
Find the smallest number that is a multiple of all the numbers from 1 to 10.
Answer

This is the previous answer with the missing 7 put back.

Smallest number = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 360 × 7 = 2520

Check:

Divide 2520 by12345678910
Answer25201260840630504420360315280252
Why we multiply only by 7: 360 already contains 2 three times, 3 twice and 5 once — enough for 1, 2, 3, 4, 5, 6, 8, 9 and 10. Only the prime 7 was missing, so multiplying by 7 once is enough. Nothing smaller can work, because the answer must contain 8 = 2 × 2 × 2, 9 = 3 × 3, 5 and 7.
Did you know? 2520 is the smallest number divisible by every number from 1 to 10. The smallest number divisible by every number from 1 to 20 is 232792560!
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