Question:

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

(a) 6

(b) 0

(c) 4

(d) 3

Solution:

(d) 3

When a number is divisible by 3, the sum of the digits must also be divisible by 3.

$x+7+y+5=(x+y)+12$

This sum is divisible by 3 if  x+y+12 is 12 or 15.

For 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

or x+y = 15-12=3