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 $a+b+c=\boxed{12}$.

See Also

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