Difference between revisions of "2003 AIME I Problems/Problem 9"

Revision as of 18:58, 4 July 2013

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 (since zero may not be the first digit). 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 $\sum_{n = 10}^{18} (19 - n)^2 = \sum_{n = 1}^9 n^2 = 285$ balanced numbers. Thus, there are in total $330 + 285 = \boxed{615}$ balanced numbers.

Both summations may be calculated using the formula for the sum of consecutive squares, namely $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ .

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Difference between revisions of "2003 AIME I Problems/Problem 9"

Revision as of 18:58, 4 July 2013

Problem

Solution

See also