1999 AIME Problems/Problem 5

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Problem

For any positive integer $\displaystyle x_{}$, let $\displaystyle S(x)$ be the sum of the digits of $\displaystyle x_{}$, and let $\displaystyle T(x)$ be $\displaystyle |S(x+2)-S(x)|.$ For example, $\displaystyle T(199)=|S(201)-S(199)|=|3-19|=16.$ How many values $\displaystyle T(x)$ 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.

See also

1999 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions