2005 Alabama ARML TST Problems/Problem 4

Problem

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

Solution

A number is divisible by $12$ if it is divisible by $3$ and $4$ . A number is divisible by $4$ if its last two digits are divisible by $4$ , so $4|\overline{B8} \Longrightarrow B = 0,2,4,6,8$ . A number is divisible by $3$ if the sum of its digits is $3$ , so

$3| s(2A08) = 10 + A \Longrightarrow A = 2,5,8$ - 3 solutions
$3| s(2A28) = 12 + A \Longrightarrow A = 0,3,6,9$ - 4 solutions
$3| s(2A48) = 14 + A \Longrightarrow A = 1,4,7$ - 3 solutions
$3| s(2A68) = 16 + A \Longrightarrow A = 2,5,8$ - 3 solutions
$3| s(2A88) = 18 + A \Longrightarrow A = 0,3,6,9$ - 4 solutions

These sum to $17$ .

2005 Alabama ARML TST (Problems)
Preceded by: Problem 3	Followed by: Problem 5
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2005 Alabama ARML TST Problems/Problem 4

Problem

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