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Class 6 Divisibility Rules — practice questions with answers

Playing with Numbers · 5 sample questions from the school syllabus. Try each one, then open the answer and the worked solution.

1. Which of these numbers is divisible by 2?
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Answer: C) 84
A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8). 84 ends in 4, so it is divisible by 2.
2. Which of these numbers is divisible by both 3 and 5?
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Answer: A) 120
Divisible by 5 means it ends in 0 or 5. Among these, 120 ends in 0 and 1 + 2 + 0 = 3 is divisible by 3. So 120 is divisible by both 3 and 5.
3. What is the smallest digit that can replace * in 2,3*5 so that the number is divisible by 9?
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Answer: D) 8
Digit sum = 2 + 3 + * + 5 = 10 + *. For divisibility by 9, this must be a multiple of 9. The next multiple of 9 from 10 is 18, so * = 18 − 10 = 8. (2,385 = 9 × 265.)
4. Rohit has the digits 2, 4 and 5, and uses each once to make a three-digit number.
He wants the number to be divisible by 5.
How many such three-digit numbers can he make?
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Answer: D) 2
To be divisible by 5, the units digit must be 5 (0 is not available). So the last digit is fixed as 5, and the remaining digits 2 and 4 fill the first two places in 2 × 1 = 2 ways: 245 and 425. So there are 2 numbers.
5. A whole number n has the property that both n and the number formed by reversing its digits are divisible by 9.
Mohit says: 'If n is divisible by 9, its digit-reversal is also divisible by 9.'
Is his statement always true? Enter 1 for Yes, 0 for No.
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Answer: C) 1
Reversing the digits does not change the digit sum (the same digits are just rearranged). Since divisibility by 9 depends only on the digit sum, the reversed number has the same digit sum and is therefore also divisible by 9. So the statement is always true: answer 1 (Yes).
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