If the 4-digit number x27y is exactly divisible by 9, then the least v

1.	If the 4-digit number x27y is exactly divisible by 9, then the least value of (x + y) is A. 0 B. 3 C. 6 D. 9
Answer» We know that if sum of digits of a number is divisible by 9, then the number is divisible by 9. Here, x + 2 + 7 + y = multiple of 9 x + y + 9= 0, 9, 18, … Hence, x + y + 9 = 9 ∴ 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 + 9 = 18 x + y = 9 ∴ the least value of (x + y) is 9

If the 4-digit number x27y is exactly divisible by 9, then the least value of (x + y) is A. 0 B. 3 C. 6 D. 9

Answer»

We know that if sum of digits of a number is divisible by 9, then the number is divisible by 9.

Here,

x + 2 + 7 + y = multiple of 9

x + y + 9= 0, 9, 18, …

Hence,

x + y + 9 = 9

∴ 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 + 9 = 18 x + y = 9

∴ the least value of (x + y) is 9