Question: Suppose x is a positive integer with n digits, say x = dd2d3 dn. In other words, di {0, 1, 2, ..., 9} for

Suppose x is a positive integer with n digits, say x =

Suppose x is a positive integer with n digits, say x = dd2d3 dn. In other words, di {0, 1, 2, ..., 9} for 1 i n, but d #0. Here's a definition: For a, b Z, a is a divisor of b if b = ak, for some k Z. Please prove the following statement: If 9 is a divisor of d + d + + dn, then 9 is a divisor of x.

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Answer Solution 2 is x ddd3 Also p p d 10n1 d we 9 10m 1 and 10 divides 9 9 is div... View full answer

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