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 1
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 .
Solution 2 (Painful Casework)
Call the number . Then . Set .
Clearly, .
If : is not acceptable.
If : The only case is or . 2 choices.
If : then since , or . There are 3 choices for : . here.
If : Clearly, because if so, the sum will be even, not odd. Counting , we have choices. Subtracting that, we have choices. Since it doesn't matter whether or , we have 4 choices for . So here.
If : Continue as above. choices for . choices for . here.
If : You get the point. .
If : .
If : .
If : .
If : .
Now we need to be careful because if , is not valid. However, we don't have to worry about .
If : . Same thing for . .
If : We start at . So .
Continue this pattern until . Add everything up: we have .
~hastapasta
See also
2003 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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