Suppose x is a positive integer with n digits, say x = dd2d3 dn. In other...

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Suppose x is a positive integer with n digits, say x = d₁d2d3 dn. In other words, di € {0, 1, 2, ..., 9} for 1 ≤ i ≤n, but d₁ #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 d₁ + d₂ + + dn, then 9 is a divisor of x. Suppose x is a positive integer with n digits, say x = d₁d2d3 dn. In other words, di € {0, 1, 2, ..., 9} for 1 ≤ i ≤n, but d₁ #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 d₁ + d₂ + + 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 the full answer