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

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==Problem==
 
==Problem==
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For how many [[ordered pair]]s of [[digit]]s <math>\displaystyle (A,B)</math> is <math>\displaystyle 2AB8</math> a [[multiple]] of 12?
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For how many [[ordered pair]]s of [[digit]]s <math>(A,B)</math> is <math>2AB8</math> a [[multiple]] of 12?
 
==Solution==
 
==Solution==
  
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{{wikify}}
 
{{wikify}}
 
==See also==
 
==See also==
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*[[2005 Alabama ARML TST]]
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{{ARML box|year=2005|state=Alabama|num-b=3|num-a=5}}
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*[[2005 Alabama ARML TST Problems/Problem 3 | Previous Problem]]
 
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*[[2005 Alabama ARML TST Problems/Problem 5 | Next Problem]]
 

Revision as of 12:41, 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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