2008 iTest Problems/Problem 47

Problem

Find $a + b + c$ , where $a$ , $b$ , and $c$ are the hundreds, tens, and units digits of the six-digit integer $123abc$ , which is a multiple of $990$ .

Solution

Because $990 = 10 \cdot 9 \cdot 11$ , the integer $123abc$ is a multiple of $10$ , $9$ , and $11$ . That means $c = 0$ , $6 + a + b \equiv 0 \pmod{9}$ , and $2 - a + b \equiv 0 \pmod{11}$ . The last two congruences can be rewritten as $a + b \equiv 3 \pmod{9}$ and $a - b \equiv 2 \pmod{11}$ . After testing values of $a$ and $b$ , we see that $a = 7$ and $b = 5$ , so $a+b+c=\boxed{12}$ .

2008 iTest (Problems)
Preceded by: Problem 46	Followed by: Problem 48
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2008 iTest Problems/Problem 47

Problem

Solution

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