2005 Alabama ARML TST Problems/Problem 4

Revision as of 11:45, 11 December 2007 by Temperal (talk | contribs) (Solution: divide latex)

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15