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

Solution 1

Denote the first of each of the series of consecutive integers as $a,\ b,\ c$ . Therefore, $n = a + (a + 1) \ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55$ . Simplifying, $9a = 10b + 9 = 11c + 19$ . 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 $10(45) + 45 = 495$ .

Solution 2

Let the desired integer be $n$ . From the information given, it can be determined that, for positive integers $a, \ b, \ c$ :

$n = 9a + 36 = 10b + 45 = 11c + 55$

This can be rewritten as the following congruences:

$n \equiv 0 \pmod{9}$

$n \equiv 5 \pmod{10}$

$n \equiv 0 \pmod{11}$

Since 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is $\boxed{495}$

Solution 3

Let $n$ be the desired integer. From the given information, we have $\begin{align*}9x &= a \\ 11y &= a \\ 10z + 5 &= a, \end{align*}$ here, $x,$ and $y$ are the middle terms of the sequence of 9 and 11 numbers, respectively. Similarly, we have $z$ as the 4th term of the sequence. Since, $a$ is a multiple of $9$ and $11,$ it is also a multiple of $\text{lcm}[9,11]=99.$ Hence, $a=99m,$ for some $m.$ So, we have $10z + 5 = 99m.$ It follows that $99(5) = \boxed{495}$ is the smallest integer that can be represented in such a way.

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
All AIME Problems and Solutions

Page

Toolbox

Search