1993 AIME Problems/Problem 6

Problem

What is the smallest positive integer than can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?

Solution

Denote the first of each of the series of consecutive integers as $a,\ b,\ c$ . Therefore, $\displaystyle n = a + (a + 1) \ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55$ . Simplifying, $\displaystyle 9a = 10b + 9 = 11c + 19$ . The relationship between $a,\ b$ suggests that $b$ is divisible by $9$ . Also, $10b -10 = 10(b-1) = 11c$ , so $b-1$ is divisible by $11$ . We find that the least possible value of $b = 45$ , so the answer is $10(45) + 45 = 495$ .

1993 AIME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1993 AIME Problems/Problem 6

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