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?

Solutions

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.

Solution 4

By the method in Solution 1, we find that the number $n$ 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.

Solution 5

First note that the integer clearly must be divisible by $9$ and $11$ since we can use the "let the middle number be x" trick. Let the number be $99k$ for some integer $k.$ Now let the $10$ numbers be $x,x+1, \cdots x+9.$ We have $10x+45 = 99k.$ Taking mod $5$ yields $k \equiv 0 \pmod{5}.$ Since $k$ is positive, we take $k=5$ thus obtaining $99 \cdot 5 = \boxed{495}$ as our answer.

Solution 6

From the first equation solution 1, we get the three part equation 9x=10y+9=11z+19 after subtracting 36 from each of the equations 9x+36=10y+45=11z+55 knowing that the integer we are looking for has a last digit of 9 from the second part of our first equation, and also knowing that our mystery number must be 19 more the a multiple of 11 that ends in 0, the only way this could happen is if the number is able to be deduced in the form 110a+19, where a is between the range 1-9, inclusive. so, also knowing our mystery number is divisible by 9, our only answer being 129,239,349,459 ... etc. until 899. Brushing through our act of casework for this problem shows that the answer is 459... but we are not done yet! We have to add the 36 we subtracted at the beggining of our solution to get the equation 459+36, which is equal to our final answer... 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
All AIME Problems and Solutions

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