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Difference between revisions of "2005 Alabama ARML TST Problems/Problem 4"

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==See also==
 
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*[[2005 Alabama ARML TST]]
 
*[[2005 Alabama ARML TST]]
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*[[2005 Alabama ARML TST/Problem 3 | Previous Problem]]
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*[[2005 Alabama ARML TST Problems/Problem 3 | Previous Problem]]
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*[[2005 Alabama ARML TST/Problem 5 | Next Problem]]
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*[[2005 Alabama ARML TST Problems/Problem 5 | Next Problem]]

Revision as of 18:31, 17 November 2006

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

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