Question
If a number 2376y32x is divisible by 9, then how many
values possible for x + y, given that x + ySolution
Given that 2376y32x is completely divisibly by 9, The divisibility rule of 9 is sum of digit should be divisible by 9Β Here 2+3+7+6+y+3+2+x, so we have here 23+x+y=should be multiple of nine. x + y= 4 on put we get 27 which is multiple of 9Β x + y=13 on put we get 36 which is multiple of 9 so here we have only 2 values possible for x + y.Β
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