# 1999 AIME Problems/Problem 5

## Problem

For any positive integer $x_{}$, let $S(x)$ be the sum of the digits of $x_{}$, and let $T(x)$ be $|S(x+2)-S(x)|.$ For example, $T(199)=|S(201)-S(199)|=|3-19|=16.$ How many values of $T(x)$ do not exceed 1999?

## Solution

For most values of $x$, $T(x)$ will equal $2$. For those that don't, the difference must be bumping the number up a ten, a hundred, etc. If we take $T(a999)$ as an example, $$|(a + 1) + 0 + 0 + 1 - (a + 9 + 9 + 9)| = |2 - 9(3)|$$ And in general, the values of $T(x)$ will then be in the form of $|2 - 9n| = 9n - 2$. From $7$ to $1999$, there are $\left\lceil \frac{1999 - 7}{9}\right\rceil = 222$ solutions; including $2$ and there are a total of $\boxed{223}$ solutions.

## See also

 1999 AIME (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 6 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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