NCERT Solutions Ganita Prakash (Part 1) Chapter 5 .1 Is This a Multiple Of? — Figure it Out

Book page 1225 Updated on2026-09-05

Q1.

The sum of four consecutive numbers is 34. What are these numbers?

Answer

The numbers are 7, 8, 9 and 10.

Let the numbers be n, n + 1, n + 2, n + 3
n + (n + 1) + (n + 2) + (n + 3) = 34
4n + 6 = 34
4n = 28
n = 7

Check: 7 + 8 + 9 + 10 = 34 ✓

Tip: 4n + 6 is always even, so four consecutive numbers can never add to an odd total. Try 35 and you will get no answer at all.

Q2.

Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.

Answer

The other four are p – 1, p – 2, p – 3 and p – 4.

In increasing order the five numbers are
p – 4, p – 3, p – 2, p – 1, p

Their sum is (p – 4) + (p – 3) + (p – 2) + (p – 1) + p = 5p – 10 = 5(p – 2), so the sum of any five consecutive numbers is a multiple of 5.

Check it yourself: take p = 20. The numbers are 16, 17, 18, 19, 20 and their sum is 90 = 5 × 18 = 5(20 – 2) ✓

Q3.

For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra. (i) The sum of two even numbers is a multiple of 3. (ii) If a number is not divisible by 18, then it is also not divisible by 9. (iii) If two numbers are not divisible by 6, then their sum is not divisible by 6. (iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3. (v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.

Answer

(i) Sometimes true.

2m + 2n = 2(m + n) — a multiple of 3 only when 3 divides m + n

Example: 2 + 4 = 6 ✓ and 4 + 8 = 12 ✓. Non-example: 2 + 6 = 8 ✗ and 6 + 8 = 14 ✗.

(ii) Sometimes true.

Example: 20 is not divisible by 18, and it is not divisible by 9 either ✓. Non-example: 27 is not divisible by 18, yet 27 = 9 × 3 is divisible by 9 ✗.

Why it happens: the multiples of 18 are every second multiple of 9 (9, 18, 27, 36, 45, 54 …). So the odd multiples of 9 — 9, 27, 45, 63 — all break this statement. What is always true is the other direction: if a number is not divisible by 9, it cannot be divisible by 18.

(iii) Sometimes true.

Example: 8 and 10 are not divisible by 6, and 8 + 10 = 18 is divisible by 6 — so the statement fails here. Non-example (where it holds): 8 and 9 are not divisible by 6, and 8 + 9 = 17 is not divisible by 6.

Write the two numbers as 6a + r and 6b + s, with r, s ≠ 0
Sum = 6(a + b) + (r + s)
Divisible by 6 exactly when r + s = 6 (or 12)

So 5 and 7 (remainders 5 and 1) give 12 ✓, while 5 and 8 (remainders 5 and 2) give 13 ✗.

(iv) Always true.

6x + 9y = 3(2x + 3y) — 3 is a factor of the whole sum

Examples: 12 + 9 = 21 = 3 × 7 ✓; 18 + 27 = 45 = 3 × 15 ✓; 6 + 90 = 96 = 3 × 32 ✓.

Why it happens: 3 divides 6 and 3 divides 9, so 3 divides both numbers — and a divisor of two numbers always divides their sum.

(v) Sometimes true.

6x + 3y = 3(2x + y) — a multiple of 9 only when 3 divides 2x + y

Example: 18 + 9 = 27 = 9 × 3 ✓. Non-example: 12 + 9 = 21 ✗ and 6 + 6 = 12 ✗.

Q4.

Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.

Answer

Such numbers are 2, 14, 26, 38, 50, 62, … and they are exactly the numbers of the form 12n + 2.

N leaves remainder 2 on ÷ 3 → N – 2 is a multiple of 3
N leaves remainder 2 on ÷ 4 → N – 2 is a multiple of 4
So N – 2 is a common multiple of 3 and 4
LCM (3, 4) = 12
N – 2 = 12n → N = 12n + 2

N	N ÷ 3	N ÷ 4
14	4 remainder 2 ✓	3 remainder 2 ✓
26	8 remainder 2 ✓	6 remainder 2 ✓
38	12 remainder 2 ✓	9 remainder 2 ✓

Why it happens: the two conditions say the same thing about N – 2 — it must be a multiple of 3 and a multiple of 4. By the rule proved on page 121, it must then be a multiple of LCM (3, 4) = 12. The list steps by 12 each time.

Q5.

“I hold some pebbles, not too many, When I group them in 3’s, one stays with me. Try pairing them up — it simply won’t do, A stubborn odd pebble remains in my view. Group them by 5, yet one’s still around, But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?”

Answer

There are 91 pebbles.

First turn each line into a condition:

Line of the riddle	Condition
groups of 3, one stays	remainder 1 on ÷ 3
pairing won't do, an odd pebble remains	remainder 1 on ÷ 2
group by 5, one's still around	remainder 1 on ÷ 5
grouping by seven, perfection	remainder 0 on ÷ 7
more than one hundred is too bold	less than 100

Remainder 1 on ÷ 2, ÷ 3 and ÷ 5
→ N – 1 is a common multiple of 2, 3 and 5
LCM (2, 3, 5) = 30
N = 30k + 1 → 1, 31, 61, 91, 121, …
Below 100 and divisible by 7: 91 = 7 × 13 ✓

Check all five lines against 91: 91 = 3 × 30 + 1 ✓; 91 is odd ✓; 91 = 5 × 18 + 1 ✓; 91 = 7 × 13 exactly ✓; 91 < 100 ✓.

Why the shortcut works: instead of testing every number below 100, notice that the first three lines all say the same thing about N – 1. That collapses three conditions into one — and only four candidates survive.

Q6.

Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true?

Answer

Yes, Tathagat's claim is true.

The three numbers are 6a + 2, 6b + 2 and 6c + 2
Sum = 6a + 6b + 6c + 6
= 6(a + b + c + 1) — a multiple of 6

Why it happens: each number carries a leftover of 2, and the three leftovers add to 6 — exactly one more complete group. So no remainder survives.

Three numbers	Sum	÷ 6
2 + 8 + 14	24	4 ✓
8 + 20 + 32	60	10 ✓
2 + 2 + 2	6	1 ✓

Try This: what about adding just two such numbers? (6a + 2) + (6b + 2) = 6(a + b) + 4 — never a multiple of 6. Three is exactly the right count.

Q7.

When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually. (i) 4779 + 661 (ii) 4779 – 661

Answer

(i) remainder 1 (ii) remainder 2.

Algebraically. Write each number as a multiple of 7 plus its remainder:

4779 = 7p + 5 661 = 7q + 3

(i) 4779 + 661 = 7p + 7q + 8
= 7(p + q) + 7 + 1
= 7(p + q + 1) + 1 → remainder 1

(ii) 4779 – 661 = 7p + 5 – 7q – 3
= 7(p – q) + 2 → remainder 2

Visually. Think of each number as counters arranged in rows of 7, with the loose ones at the bottom.

Only the loose counters decide the remainder; the full rows cancel out.

Check it yourself: 4779 + 661 = 5440 = 7 × 777 + 1 ✓ and 4779 – 661 = 4118 = 7 × 588 + 2 ✓ — but no division was needed to say so.

Q8.

Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?

Answer

The smallest such number is 59.

Look at the three conditions side by side: each remainder is exactly 1 less than its divisor.

Divisor	Remainder	Meaning
3	2	1 short of a multiple of 3
4	3	1 short of a multiple of 4
5	4	1 short of a multiple of 5

So N + 1 is a multiple of 3, of 4 and of 5
N + 1 is a multiple of LCM (3, 4, 5) = 3 × 4 × 5 = 60
Smallest such N + 1 is 60 → N = 59

Check: 59 = 3 × 19 + 2 ✓, 59 = 4 × 14 + 3 ✓, 59 = 5 × 11 + 4 ✓.

Why 59 is the smallest: N + 1 must be a common multiple of 3, 4 and 5, and 60 is the smallest positive common multiple there is. Any smaller candidate would force N + 1 to be a smaller common multiple, which does not exist. The next numbers in the family are 119, 179, 239 — each 60 more than the last.

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