Playing With Numbers - Class 8 Mathematics - MCQ Test 3 - Class 8 MCQ

The last digit is 5, the number is divisible by 5. Since 311636−10 = 311626 ÷ 7 = 44568, the number is divisible by 7 also. 

A number divisible by 3 and 4 is divisible by 12.    

Adding the digits at odd places of the given number 653 ∗ 47,we get 13.
Now if the number is divisible by 11, the sum of digits at even places should also be 13, so that 13−13  = 0 is divisible by 11. We have 7+∗+5 = 13.
∴  ∗ = 13 −12 = 1is the required value.

1001 is divisible by 13, as 1001 = 13×77.

The prime number after 32 is 37 which is got by adding 5 to 32.

153 = 3×51. So, 153 is not prime.

147 = 7×21 is a composite number.

The prime numbers between 30 and 60 are 31, 37, 41, 43, 47, 53 and 59. The greatest of these is 59.

The statement in option (D) is true according to the test of divisibility by 8.

According to the problem, a five-digit number between 10000 and 10010 divisible by 5 is 10005.

672 ÷ 8 = 84
∴ 505672 is a multiple of 8.

The prime numbers between 100 and 200 are 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197 and 199 which are 21 in number.