Question: Let n be a positive integer with base-10 representation d1 . . . dn , where each di is a digit in the set {0,

Let n be a positive integer with base-10 representation d1 . . . dn , where each di is a digit in the set {0, 1, . . . , 9}. Prove that n mod 9 = X i di mod 9 . Amazing, right? For example 3964 mod 9 = (3 9 6 4) mod 9 = 4 . (Hint: What is 10k mod 9?) Back story: This rather nice fact about reduction modulo 9 leads to an old accounting trick. Suppose that you want to check that the result of a (big, nasty) multiplication (say) is correct, e.g.: d1 . . . dk e1 . . . e = f1 . . . fm (where each di , ei , and fi is a base-10 digit). One can check that it is correct modulo 9, which is to say that the two sides are the same when they are reduced mod 9. Now, it is very easy to take each side modulo 9; on the left side, this can be computed by simply adding up the digits themselves modulo 9, taking the product modulo 9, resumming the digits modulo 9, and using the rule again. For example

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