Problem: Let f(n) be the sum of n and its digits. For example, f(34)=41. Is there an integer such that f(n)=1980? Show that given any positive integer m we can find n such that f(n)=m or m+1.
Solution
Solution: If the last digit of n is not 9, then f(n+1)=f(n)+2. If the last digit of n is 9, then f(n+1)<f(n). On the other hand, f clearly achieves arbitrarily large values. Also f(1)=1.
Now consider any m>1. Let M be the smallest integer such that f(M)>m. Then f(M−1)≤m. Since f(M)>f(M−1) we must have f(M)=f(M−1)+2. Hence either f(M)=m+1 or f(M−1)=m.
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