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 (since zero may not be the first digit). 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}^9 n^2 = 285</math> balanced numbers. Thus, there are in total <math>330 + 285 = \boxed{615}</math> balanced numbers. | + | If the common sum of the first two and last two digits is <math>n</math>, such that <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 (since zero may not be the first digit). 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>, such that <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}^9 n^2 = 285</math> balanced numbers. Thus, there are in total <math>330 + 285 = \boxed{615}</math> balanced numbers. |
| − | Both summations may be calculated using the formula for the [[perfect square|sum of consecutive squares]], namely <math>\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}</math>. | + | Both summations may be calculated using the formula for the [[perfect square|sum of consecutive squares]], namely <math>\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}</math>. |
== See also == | == See also == | ||
Revision as of 16:39, 9 August 2018
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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
