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NCERT Solutions Ganita Prakash Chapter 5 & 126Section 5.5 Divisibility Tests — Figure it Out

Book page 125 Updated on2026-09-05

Q1.
2024 is a leap year (as February has 29 days). Leap years occur in the years that are multiples of 4, except for those years that are evenly divisible by 100 but not 400. a. From the year you were born till now, which years were leap years? b. From the year 2024 till 2099, how many leap years are there?
Answer

a. Write down your birth year and keep adding until you reach a multiple of 4; after that every 4th year is a leap year (there is no “century” exception anywhere between 2001 and 2099).

Example: if you were born in 2013, the leap years since then are
2016, 2020, 2024 — and the next ones will be 2028, 2032, 2036, …
If you were born in 2014 → 2016, 2020, 2024
If you were born in 2012 → 2012, 2016, 2020, 2024

Test any year quickly with the rule for 4 — look at the last two digits: 16 ÷ 4 = 4 ✔, 20 ÷ 4 = 5 ✔, 24 ÷ 4 = 6 ✔, but 22 ÷ 4 leaves 2 ✘.

b. From 2024 till 2099 there are 19 leap years.

They are 2024, 2028, 2032, …, 2096
Count = (2096 − 2024) ÷ 4 + 1 = 72 ÷ 4 + 1 = 18 + 1 = 19
Why the “100 but not 400” rule does not bite here: the only multiple of 100 nearby is 2100, and it lies outside our range. (For the record, 2100 will not be a leap year, because it is divisible by 100 but not by 400 — while 2000 was a leap year, since 2000 ÷ 400 = 5.)
Q2.
Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes.
Answer

A 4-digit palindrome looks like abba — the first and last digits are the same, and so are the middle two. For it to be divisible by 4, the number made by the last two digits, ba, must be divisible by 4. Since a multiple of 4 is always even, the digit a must be even (2, 4, 6 or 8 — it cannot be 0, as a 4-digit number cannot start with 0).

Largest: take a = 8 and make b as large as possible.

Test b8 for b = 9, 8, 7, …
98 ÷ 4 → remainder 2 ✘    88 ÷ 4 = 22 ✔
So the number is 8888, and 8888 ÷ 4 = 2222 ✔

Smallest: take a = 2 and make b as small as possible.

Test b2 for b = 0, 1, 2, …
02 ÷ 4 → remainder 2 ✘    12 ÷ 4 = 3 ✔
So the number is 2112, and 2112 ÷ 4 = 528 ✔

Answer: largest = 8888, smallest = 2112.

Note — the book's answer key is wrong here. It gives 9999 and 1001, but both of these are odd numbers, so neither can be divisible by 4 (9999 ÷ 4 = 2499 remainder 3; 1001 ÷ 4 = 250 remainder 1). The correct answers are 8888 and 2112.
Q3.
Explore and find out if each statement is always true, sometimes true or never true. You can give examples to support your reasoning. a. Sum of two even numbers gives a multiple of 4. b. Sum of two odd numbers gives a multiple of 4.
Answer

a. Sometimes true.

2 + 6 = 8 = 4 × 2 ✔    10 + 14 = 24 = 4 × 6 ✔
but 2 + 4 = 6 ✘    8 + 10 = 18 ✘

b. Sometimes true.

1 + 3 = 4 = 4 × 1 ✔    7 + 9 = 16 = 4 × 4 ✔
but 1 + 5 = 6 ✘    5 + 9 = 14 ✘
When does it work? The sum of two even numbers is always even, and the sum of two odd numbers is also always even — so in both cases we get a multiple of 2, but not always of 4.
  • Two even numbers give a multiple of 4 when both leave the same remainder on division by 4 (both like 2, 6, 10, 14 …, or both like 4, 8, 12, 16 …).
  • Two odd numbers give a multiple of 4 when one is of the form 4k + 1 (1, 5, 9, 13 …) and the other is of the form 4k + 3 (3, 7, 11, 15 …).
So neither statement is always true and neither is never true — both are sometimes true.
Q4.
Find the remainders obtained when each of the following numbers are divided by (a) 10, (b) 5, (c) 2. 78, 99, 173, 572, 980, 1111, 2345
Answer

You do not need long division at all — just look at the last digit.

  • By 10: the remainder is the units digit.
  • By 5: take the units digit and subtract 5 if it is 5 or more.
  • By 2: the remainder is 0 for an even units digit and 1 for an odd one.
NumberUnits digitRemainder ÷ 10Remainder ÷ 5Remainder ÷ 2
788830
999941
1733331
5722220
9800000
11111111
23455501
Sample checks: 78 = 7 × 10 + 8    78 = 15 × 5 + 3    78 = 39 × 2 + 0
2345 = 234 × 10 + 5    2345 = 469 × 5 + 0    2345 = 1172 × 2 + 1
Q5.
The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked for divisibility of 14560 by only two of these numbers and then declared that it was also divisible by all of them. What could those two numbers be?
Answer

The two numbers are 8 and 5.

Last three digits of 14560 = 560, and 560 ÷ 8 = 70 → divisible by 8 ✔
Units digit is 0 → divisible by 5 ✔

Those two checks give the other three for free:

  • Divisible by 8 → also divisible by 4 and by 2, since 8 = 4 × 2 (14560 ÷ 8 = 1820, so 14560 = 4 × 3640 = 2 × 7280).
  • Divisible by 2 and by 5 → divisible by 10, since 10 = 2 × 5 (14560 ÷ 10 = 1456).
14560 ÷ 2 = 7280   14560 ÷ 4 = 3640   14560 ÷ 5 = 2912   14560 ÷ 8 = 1820   14560 ÷ 10 = 1456 ✔
Why 8 and 5 are the clever choice: 8 is the “biggest” of the powers of 2 in the list, so it covers 4 and 2 below it. And 5 together with the 2 (that came free from 8) covers 10. In prime language, 14560 = 2 × 2 × 2 × 2 × 2 × 5 × 7 × 13 — the check for 8 and 5 confirms that 2 × 2 × 2 and 5 are present, which is all the other three tests need.
Q6.
Which of the following numbers are divisible by all of 2, 4, 5, 8 and 10: 572, 2352, 5600, 6000, 77622160.
Answer

Being divisible by 2, 4, 5, 8 and 10 all together is the same as being divisible by 40 (= 8 × 5). So check two things: last three digits divisible by 8, and units digit 0 or 5 — in fact 0, since the number must also be even.

Number÷ 5 and ÷ 10?Last three digits ÷ 8?Divisible by all five?
572ends in 2 → No572 ÷ 8 leaves 4No
2352ends in 2 → No352 ÷ 8 = 44 ✔No
5600ends in 0 → Yes600 ÷ 8 = 75 ✔Yes ✔
6000ends in 0 → Yes000 → 0 ÷ 8 = 0 ✔Yes ✔
77622160ends in 0 → Yes160 ÷ 8 = 20 ✔Yes ✔

Answer: 5600, 6000 and 77622160.

5600 ÷ 40 = 140    6000 ÷ 40 = 150    77622160 ÷ 40 = 1940554 ✔
Careful with 2352: it passes the tests for 2, 4 and 8, but it fails for 5 and 10 — so it does not qualify. And 572 passes only for 2 and 4 (572 ÷ 4 = 143).
Q7.
Write two numbers whose product is 10000. The two numbers should not have 0 as the units digit.
Answer

Answer: 16 and 625.

16 × 625 = 10000 ✔ — neither number ends in 0

Here is how to find it. Start with the prime factorisation:

10000 = 10 × 10 × 10 × 10 = (2 × 5) × (2 × 5) × (2 × 5) × (2 × 5)
10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

A number ends in 0 only when it has both a 2 and a 5 among its factors. So put all the 2s in one number and all the 5s in the other:

2 × 2 × 2 × 2 = 16   and   5 × 5 × 5 × 5 = 625
Why this is the only choice: the four 2s and four 5s have to be shared out between the two numbers. If any number gets a 2 and a 5, it becomes a multiple of 10 and ends in 0. So one number must take all the 2s and the other all the 5s — giving 16 × 625 and nothing else.
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