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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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{{ARML box|year=2005|state=Alabama|num-b=3|num-a=5}}
 
{{ARML box|year=2005|state=Alabama|num-b=3|num-a=5}}
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[[Category:Intermediate Number Theory Problems]]

Revision as of 12:44, 11 December 2007

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

$A+2C=2\rightarrow$ 4

$A+2C=2\rightarrow$ 4

$A+2C=2\rightarrow$ 3

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

Total of 18 ways.

Template:Wikify

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
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