2003 AIME I Problems/Problem 9
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 .
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 .
We ignore the requirement that the first digit is non-zero, and do casework on the sum of the sum of the pairs of digits.
If two digits and sum to , we have possibility:
If , we have possibilities: and
: , , and are the only possibilities.
We notice a pattern: for all , there are ordered pairs of digits such that . Then, testing for , we notice that there are ordered pairs with a sum of .
So the number of ways to pick digits satisfying the constraint is
We have to subtract out the cases where the first digit is , which is
So our answer is
This solution takes some inspiration from dice. In a way, we are counting the number of ways to roll a given number with a pair of -sided dice.
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