If x7y5 is exactly divisible by 3, then the least value of (x + y) is�

1.	If x7y5 is exactly divisible by 3, then the least value of (x + y) is A. 6 B. 0 C. 4 D. 3
Answer» We know that if sum of digits of a number is divisible by 3, then the number is divisible by 3. Here, x + 7 + y + 5 = multiple of 3 x + y + 12 = 0, 3, 6, 9, …. Hence, x + y + 12 = 12 ∴ x + y = 0 But x + y cannot be 0 because then x and y both will have to be 0.Since x is the first digit, it cannot be 0. ∴ x + y + 12 = 15 x + y = 3 ∴ the least value of (x + y) is 3

If x7y5 is exactly divisible by 3, then the least value of (x + y) is A. 6 B. 0 C. 4 D. 3

Answer»

We know that if sum of digits of a number is divisible by 3, then the number is divisible by 3.

Here,

x + 7 + y + 5 = multiple of 3

x + y + 12 = 0, 3, 6, 9, ….

Hence,

x + y + 12 = 12

∴ x + y = 0

But x + y cannot be 0 because then x and y both will have to be 0.Since x is the first digit, it cannot be 0.

∴ x + y + 12 = 15 x + y = 3

∴ the least value of (x + y) is 3