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NCERT Solutions Ganita Prakash Chapter 5 In-text Questions — Divisibility Tests

Book page 123 to 125 Updated on2026-09-05

Q1.
The first few multiples of 10 are: 10, 20, 30, 40, … Is 125 a multiple of 10? Will this number appear in the previous sequence? Why or why not? Can you now answer if 8560 is divisible by 10?
Answer

125 is not a multiple of 10, and it will never appear in that sequence.

10, 20, 30, …, 110, 120, 130, 140, … — the list jumps straight from 120 to 130
125 ÷ 10 = 12 remainder 5

Why: every multiple of 10 ends in 0, because 10 × any number puts a zero in the units place. But 125 ends in 5, so it can never be in the list.

Is 8560 divisible by 10? Yes.

8560 ends in 0 → 8560 = 856 × 10 → 8560 ÷ 10 = 856, remainder 0 ✔
Q2.
Consider this statement: Numbers that are divisible by 10 are those that end with ‘0’. Do you agree?
Answer

Yes, we fully agree. This works both ways round.

  • Every multiple of 10 ends in 0 — because multiplying by 10 shifts every digit one place to the left and drops a 0 into the units place. 7 × 10 = 70, 43 × 10 = 430.
  • Every number ending in 0 is a multiple of 10 — because you can rub off the last 0 and read the rest as the quotient. 2560 = 256 × 10.
Numbers ending in 0 → 10, 40, 90, 250, 8560 → all divisible by 10 ✔
Numbers not ending in 0 → 125, 682, 8536 → none divisible by 10 ✘
The reason behind it: any number can be split as (all digits except the last) × 10 + (units digit). The first part is already a multiple of 10, so the whole number is a multiple of 10 exactly when the units digit is 0.
Q3.
Explore by listing down the multiples: 5, 10, 15, 20, 25, ... What do you observe about these numbers? Do you see a pattern in the last digit? What is the largest number less than 399 that is divisible by 5? Is 8560 divisible by 5?
Answer

List the multiples of 5 and look at the last digit:

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, …

The last digit is always 0 or 5, turn and turn about.

Largest number less than 399 divisible by 5 = 395.

399 ÷ 5 = 79 remainder 4 → 5 × 79 = 395
(400 is also a multiple of 5, but it is bigger than 399.)

Is 8560 divisible by 5? Yes.

8560 ends in 0 → 8560 ÷ 5 = 1712, remainder 0 ✔
Q4.
Consider this statement: Numbers that are divisible by 5 are those that end with either a ‘0’ or a ‘5’. Do you agree?
Answer

Yes, we agree.

Ends in 0 or 5 → 25, 60, 195, 395, 8560 → all divisible by 5 ✔
Ends in any other digit → 78, 99, 173, 572, 8536 → none divisible by 5 ✘
Why it is true: split the number as (all digits except the last) × 10 + (units digit). Since 10 is a multiple of 5, the first part is always a multiple of 5. So the whole number is a multiple of 5 exactly when the units digit is a multiple of 5 — and among the digits 0 to 9 only 0 and 5 are multiples of 5.
Tip: notice how the same trick explains both the 10 test and the 5 test. Only the last digit matters for both, because 10 is divisible by both 10 and 5.
Q5.
The first few multiples of 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, ... What do you observe? Do you see a pattern in the last digit? Is 682 divisible by 2? Can we answer this without doing the long division? Is 8560 divisible by 2? Why or why not?
Answer

Look at the last digits of the multiples of 2:

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, …

The last digit is always one of 0, 2, 4, 6, 8, repeating in a cycle of five.

Is 682 divisible by 2? Yes — and we do not need long division.

682 ends in 2, which is even → 682 ÷ 2 = 341, remainder 0 ✔

Is 8560 divisible by 2? Yes, because it ends in 0.

8560 ÷ 2 = 4280, remainder 0 ✔
Q6.
Consider this statement: Numbers that are divisible by 2 are those that end with ‘0’, ‘2’, ‘4’, ‘6’ or ‘8’. Do you agree? What are all the multiples of 2 between 399 and 411?
Answer

Yes, we agree — these are exactly the even numbers.

Why: once again split the number as (all digits except the last) × 10 + (units digit). 10 is even, so the first part is always even. The number is therefore even exactly when its units digit is even — that is, 0, 2, 4, 6 or 8.

Multiples of 2 between 399 and 411:

400, 402, 404, 406, 408 and 410

That is six numbers — every alternate number in that stretch.

Check: 400 ÷ 2 = 200, 402 ÷ 2 = 201, 404 ÷ 2 = 202, 406 ÷ 2 = 203, 408 ÷ 2 = 204, 410 ÷ 2 = 205. ✔
Q7.
Find numbers between 330 and 340 that are divisible by 4. Also, find numbers between 1730 and 1740, and 2030 and 2040, that are divisible by 4. What do you observe? Is 8536 divisible by 4?
Answer
Stretch of numbersDivisible by 4Last two digits
Between 330 and 340332, 33632, 36
Between 1730 and 17401732, 173632, 36
Between 2030 and 20402032, 203632, 36

What we observe: in all three stretches the same two endings turn up — 32 and 36. The digits in front make no difference at all. So only the last two digits decide divisibility by 4.

Is 8536 divisible by 4? Yes.

Last two digits = 36, and 36 ÷ 4 = 9 ✔
So 8536 ÷ 4 = 2134, remainder 0 ✔
Careful: the last single digit is of no use for 4. Both 12 and 22 end in 2, but 12 is a multiple of 4 and 22 is not.
Q8.
Consider these statements: 1. Only the last two digits matter when deciding if a given number is divisible by 4. 2. If the number formed by the last two digits is divisible by 4, then the original number is divisible by 4. 3. If the original number is divisible by 4, then the number formed by the last two digits is divisible by 4. Do you agree? Why or why not?
Answer

Yes — all three statements are true.

Why: any number can be split as
Number = (the part in front) × 100 + (the last two digits)
and 100 = 4 × 25, so the first piece is always a multiple of 4, whatever the front digits may be. So the whole number is a multiple of 4 exactly when the last two digits form a multiple of 4 — which is precisely what all three statements say.
8536 = 85 × 100 + 36 = 85 × (4 × 25) + (4 × 9) = 4 × (2125 + 9) = 4 × 2134 ✔
4028 = 40 × 100 + 28 → 28 ÷ 4 = 7 ✔    364 = 3 × 100 + 64 → 64 ÷ 4 = 16 ✔
But 8542 → last two digits 42, and 42 ÷ 4 leaves 2 → 8542 is not divisible by 4 ✘
Try This: pick any four-digit number, keep the last two digits and change the first two as you like — 1132, 5532, 9932. All of them stay divisible by 4, because 32 is.
Q9.
Find numbers between 120 and 140 that are divisible by 8. Also find numbers between 1120 and 1140, and 3120 and 3140, that are divisible by 8. What do you observe? Change the last two digits of 8560 so that the resulting number is a multiple of 8.
Answer
Stretch of numbersDivisible by 8Last three digits
Between 120 and 140128, 136128, 136
Between 1120 and 11401128, 1136128, 136
Between 3120 and 31403128, 3136128, 136

What we observe: the same two endings — 128 and 136 — appear every time. The thousands digit changes nothing. So only the last three digits decide divisibility by 8.

Changing the last two digits of 8560:

8560 → last three digits 560, and 560 ÷ 8 = 70
So 8560 is already a multiple of 8 (8560 = 8 × 1070).

If we must change the last two digits, plenty of choices work — the last three digits just have to stay a multiple of 8:

8504, 8512, 8520, 8528, 8536, 8544, 8552, 8568, 8576, 8584, 8592
For example 8552 = 8 × 1069 ✔   and   8536 = 8 × 1067 ✔
Note: the answer key at the back of the book gives 8552, which is correct — but it is worth noticing that 8560 itself was already divisible by 8.
Q10.
Consider these statements: 1. Only the last three digits matter when deciding if a given number is divisible by 8. 2. If the number formed by the last three digits is divisible by 8, then the original number is divisible by 8. 3. If the original number is divisible by 8, then the number formed by the last three digits is divisible by 8. Do you agree? Why or why not?
Answer

Yes — all three statements are true.

Why: split the number as
Number = (the part in front) × 1000 + (the last three digits)
and 1000 = 8 × 125, so the first piece is always a multiple of 8. Whatever happens in front simply cannot change the answer — only the last three digits can.
8576 = 8 × 1000 + 576 → 576 ÷ 8 = 72 ✔
7648 = 7 × 1000 + 648 → 648 ÷ 8 = 81 ✔
5024 = 5 × 1000 + 024 → 24 ÷ 8 = 3 ✔
But 8570 → 570 ÷ 8 leaves remainder 2 → 8570 is not divisible by 8 ✘
See the pattern: 10 = 10 × 1 → one digit for 10 and 5; 100 = 4 × 25 → two digits for 4; 1000 = 8 × 125 → three digits for 8. Each time we double the divisor we need one more digit.
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