Difference between revisions of "2005 Alabama ARML TST Problems/Problem 4"
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We wish for <math>2000+100A+10B+8 \equiv 0 \pmod 12\rightarrow 4A+10B\equiv 8 \pmod 12\rightarrow 2A+5B\equiv 4 \pmod 6</math>. Thus <math>B\equiv 0 \pmod 2</math>. Let <math>B=2C\rightarrow A+2C\equiv 2 \pmod 3</math>; <math>C<5</math>,<math>A<10</math>, one of the eqns. must be true: | We wish for <math>2000+100A+10B+8 \equiv 0 \pmod 12\rightarrow 4A+10B\equiv 8 \pmod 12\rightarrow 2A+5B\equiv 4 \pmod 6</math>. Thus <math>B\equiv 0 \pmod 2</math>. Let <math>B=2C\rightarrow A+2C\equiv 2 \pmod 3</math>; <math>C<5</math>,<math>A<10</math>, one of the eqns. must be true: | ||
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<math>A+2C=2\rightarrow</math> 2 ways | <math>A+2C=2\rightarrow</math> 2 ways | ||
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<math>A+2C=5\rightarrow</math> 3 | <math>A+2C=5\rightarrow</math> 3 | ||
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<math>A+2C=2\rightarrow</math> 4 | <math>A+2C=2\rightarrow</math> 4 | ||
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<math>A+2C=2\rightarrow</math> 4 | <math>A+2C=2\rightarrow</math> 4 | ||
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<math>A+2C=2\rightarrow</math> 3 | <math>A+2C=2\rightarrow</math> 3 | ||
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<math>A+2C=2\rightarrow</math> 2 ways | <math>A+2C=2\rightarrow</math> 2 ways | ||
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Total of 18 ways. | Total of 18 ways. | ||
==See also== | ==See also== | ||
Revision as of 13:45, 17 November 2006
Problem
For how many ordered pairs of digits 
is 
a multiple of 12?
Solution
We wish for 
. Thus 
. Let 
; 
,
, one of the eqns. must be true:
2 ways
3
4
4
3
2 ways
Total of 18 ways.
