Problem 1

For a positive integer n, let S(n) be the sum of the digits in the decimal representation of n. A positive integer obtained by removing at least one digit from the right-hand end of n is called a stump of n. Let T(n) be the sum of all stumps of n. Prove that n=S(n)+9T(n).

Step 1 of 4: Record the digit recurrences

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T(m)=n+T(n),\qquad S(m)=S(n)+a

Detailed analysis

If n has k digits and m=10n+a is obtained by appending a digit a, then the stumps of m consist of n together with all stumps of n. Thus T(m)=n+T(n), while S(m)=S(n)+a.