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 $223$ solutions.

1999 AIME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1999 AIME Problems/Problem 5

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