2003 AIME I Problems/Problem 9

Problem

An integer between 1000 and 9999, inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers are there?

Solution

If the common sum of the first two and last two digits is $n$ , $1 \leq n \leq 9$ , there are $n$ choices for the first two digits and $n + 1$ choices for the second two digits. This gives $\sum_{n = 1}^9 n(n + 1) = 330$ balanced numbers. If the common sum of the first two and last two digits is $n$ , $10 \leq n \leq 18$ , there are $19 - n$ choices for both pairs. This gives $\sum_{n = 10}^{18} (19 - n)^2 = \sum_{n = 1}^n n^2 = 285$ balanced numbers. Thus, there are in total $330 + 285 = 615$ balanced numbers.

2003 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2003 AIME I Problems/Problem 9

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