1993 AIME Problems/Problem 6

Problem

What is the smallest positive integer that 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 $10(45) + 45 =$ \boxed{495}$.

=== Solution 2 ===

Let the desired integer be$ (Error compiling LaTeX. Unknown error_msg)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}$$ (Error compiling LaTeX. Unknown error_msg)n \equiv 5 \pmod{10}$$ (Error compiling LaTeX. Unknown error_msg)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$ (Error compiling LaTeX. Unknown error_msg)n$be the desired integer. From the given information, we have <cmath> \begin{align*}9x &= a \\ 11y &= a \\ 10z + 5 &= a, \end{align*}</cmath> here,$ (Error compiling LaTeX. Unknown error_msg)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.

=== Solution 4 === By the method in Solution 1, we find that the number$ (Error compiling LaTeX. Unknown error_msg)n $can be written as$ 9a+36=10b+45=11c+55 $for some integers$ a,b,c $. From this, we can see that$ n $must be divisible by 9, 5, and 11. This means$ n $must be divisible by 495. The only multiples of 495 that are small enough to be AIME answers are 495 and 990. From the second of the three expressions above, we can see that$ n $cannot be divisible by 10, so$ n $must equal$ \boxed{495}$. Solution by Zeroman.

1993 AIME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1993 AIME Problems/Problem 6

Contents

Problem

Solution

Solution 1

See also