2005 Alabama ARML TST Problems/Problem 4

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Problem

For how many ordered pairs of digits $(A,B)$ is $2AB8$ 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 ways

$A+2C=2\rightarrow$ 4 ways

$A+2C=2\rightarrow$ 4 ways

$A+2C=2\rightarrow$ 3 ways

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

We have a total of $18$ ways.

See also

2005 Alabama ARML TST (Problems)
Preceded by:
Problem 3
Followed by:
Problem 5
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