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}\Longleftrightarrow 4A+10B\equiv 8 \pmod {12} \Longleftrightarrow 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}\Longleftrightarrow 4A+10B</math><math>\equiv 8 \pmod {12} \Longleftrightarrow 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: |
<math>A+2C=2\rightarrow</math> 2 ways | <math>A+2C=2\rightarrow</math> 2 ways | ||
Revision as of 12:45, 11 December 2007
Problem
For how many ordered pairs of digits 
is 
a multiple of 
?
Solution
We wish for 
. Thus 
. Let 
; 
,
, one of the eqns. must be true:
2 ways
3 ways
4 ways
4 ways
3 ways
2 ways
We have a total of 
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 | ||
