2003 AIME I Problems/Problem 9
Problem
An integer between and , 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 , such that , there are choices for the first two digits and choices for the second two digits (since zero may not be the first digit). This gives balanced numbers. If the common sum of the first two and last two digits is , such that , there are choices for both pairs. This gives balanced numbers. Thus, there are in total balanced numbers.
Both summations may be calculated using the formula for the sum of consecutive squares, namely .
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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