Difference between revisions of "2003 AIME I Problems/Problem 9"
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[[Category: Intermediate Combinatorics Problems]] | [[Category: Intermediate Combinatorics Problems]] | ||
Revision as of 17:34, 8 March 2007
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 
, 
, there are choices for the first two digits and 
choices for the second two digits. This gives 
balanced numbers. If the common sum of the first two and last two digits is , 
, there are 
choices for both pairs. This gives 
balanced numbers. Thus, there are in total 
balanced numbers.
See also
| 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 | ||
| All AIME Problems and Solutions | ||
