Question
If β1x5629β is a six digit number which is divisible
by 9, then which of the following can be the minimum value of βxβ?Solution
For a number to be divisible by β9β, the sum of its digits must be divisible by β9β.So, (1 + x + 5 + 6 + 2 + 9) = (23 + x)So, number after 23 which is multiple of 9 is 27,Therefore, minimum value of βxβ = 27 β 23 = 4
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