2005 Alabama ARML TST Problems/Problem 4

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Problem

For how many ordered pairs of digits $\displaystyle (A,B)$ is $\displaystyle 2AB8$ a multiple of 12?

Solution

We wish for $2000+100A+10B+8 \equiv 0 \pmod {12}\Longleftrightarrow 4A+10B\equiv 8 \pmod {12} \Longleftrightarrow 2A+5B\equiv 4 \pmod 6$. Thus $B\equiv 0 \pmod 2$. Let $B=2C\rightarrow A+2C\equiv 2 \pmod 3$; $C<5$,$A<10$, one of the eqns. must be true:

$A+2C=2\rightarrow$ 2 ways

$A+2C=5\rightarrow$ 3

$A+2C=2\rightarrow$ 4

$A+2C=2\rightarrow$ 4

$A+2C=2\rightarrow$ 3

$A+2C=2\rightarrow$ 2 ways

Total of 18 ways.

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See also