2006 SMT/Algebra Problems/Problem 5

Problem

Jerry is bored one day, so he makes an array of Cocoa pebbles. He makes $8$ equal rows with the pebbles remaining in a box. When Kramer drops by and eats one, Jerry yells at him until Kramer realizes he can make $9$ equal rows with the remaining pebbles. After Kramer eats another, he finds he can make $10$ equal rows with the remaining pebbles. Find the smallest number of pebbles that were in the box in the beginning.

Solution

Let $n$ be the minimum number of pebbles in the box. We're given that $x$ is a multiple of $8$ , one more than a mulitple of $9$ , and two more than a multiple of $10$ .

Since $n$ is two more than a multiple of $10$ , its last digit must be $2$ . Since it's one more than a multiple of $9$ , its digits must sum to one more than a multiple of $9$ .

First, let's guess that $n$ is a two digit number. It must be of the form $\_2$ , and its digits must sum to $10$ , so $n=82$ . However, this is not a multiple of $8$ , so $n$ cannot be only two digits.

Next, we try $n=\_\_2$ , a three digit number. The sum of the digits should be $10$ . (We don't try $19$ because that would make the number too big, and we're only looking for the smallest value.) First, we try $n=172$ , but this isn't a multiple of $8$ . $n=262$ isn't a multiple of $8$ either, but $n=352$ is. Therefore, the smallest value of $n$ is $\boxed{352}$ .

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2006 SMT/Algebra Problems/Problem 5

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