Let *S*(*n*) denote the sum of the (base 10) digits of
*n*. Show that for any positive integer *m* there is an
*n* such that

For example, when *m* = 4, *n* = 13 works since 13^{2}
= 169 and *S*(169)/*S*(13) = 16/4 = 4.

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