Question

If 57x91y is divisible by 36, where x and y are digits, then the maximum value of (x + y) is​

Answer 1

Given info : 57x91y is divisible by 36 where x and y are digits.

To find : the maximum value of (x + y) is ..

solution : 57x91y is divisible by 36 it means 57x91 is divisible by 4 and 9.

divisiblity of 4 : last two digits must be divisible by 4.

57x91y is divisible by 4 only when 1y is divisible by 4. and it is possible only when y = 2 or 6 ...(1)

divisiblity of 9 : sum of all digits must be integral multiple of 9.

so, 5 + 7 + x + 9 + 1 + y = 22 + x + y is divisible by 9 when 22 + x + y = 9n , where n is integral multiple.

if y = 2,

22 + x + 2 = 9n

⇒24 + x = 9n

to get positive and one digit value of x we have to choose n = 3

so, 24 + x = 9 × 3 ⇒x = 3

so x = 3 and y = 2

similarly , if y = 6

22 + x + 6 = 9n

⇒28 + x = 9 × 4

⇒28 + x = 36

⇒x = 8

so x = 8 and y = 6

so the maximum value of (x + y) = (8 + 6) = 14