If 31y5 is a multiple of (a) 3 and (b) 9, find the value of y which satisfies the number, where y be any digit.

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if no is divisible by 9 then it will be divisible by 3 to.'so we have to put a digit which makes the sum of the digits divisible by 9.
3+1+5=9
thus if we put 0(then sum=9) or 9(sum=18) then no is divisible by 9 as 9 & 18 are divisible by 9
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