Difference between revisions of "2003 AIME I Problems/Problem 9"
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== Solution == | == Solution == | ||
| − | {{ | + | If the common sum of the first two and last two digits is <math>n</math>, <math>1 \leq n \leq 9</math>, there are <math>n</math> choices for the first two digits and <math>n + 1</math> choices for the second two digits. This gives <math>\sum_{n = 1}^9 n(n + 1) = 330</math> balanced numbers. If the common sum of the first two and last two digits is <math>n</math>, <math>10 \leq n \leq 18</math>, there are <math>19 - n</math> choices for both pairs. This gives <math>\sum_{n = 10}^{18} (19 - n)^2 = \sum_{n = 1}^n n^2 = 285</math> balanced numbers. Thus, there are in total <math>330 + 285 = 615</math> balanced numbers. |
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== See also == | == See also == | ||
* [[2003 AIME I Problems/Problem 8 | Previous problem]] | * [[2003 AIME I Problems/Problem 8 | Previous problem]] | ||
Revision as of 18:06, 19 January 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.
