Difference between revisions of "2005 Alabama ARML TST Problems/Problem 4"
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==Problem== | ==Problem== | ||
| − | For how many [[ordered pair]]s of [[digit]]s <math>(A,B)</math> is <math>2AB8</math> a [[multiple]] of 12? | + | For how many [[ordered pair]]s of [[digit]]s <math>(A,B)</math> is <math>2AB8</math> a [[multiple]] of <math>12</math>? |
==Solution== | ==Solution== | ||
| Line 7: | Line 7: | ||
<math>A+2C=2\rightarrow</math> 2 ways | <math>A+2C=2\rightarrow</math> 2 ways | ||
| − | <math>A+2C=5\rightarrow</math> 3 | + | <math>A+2C=5\rightarrow</math> 3 ways |
| − | <math>A+2C=2\rightarrow</math> 4 | + | <math>A+2C=2\rightarrow</math> 4 ways |
| − | <math>A+2C=2\rightarrow</math> 4 | + | <math>A+2C=2\rightarrow</math> 4 ways |
| − | <math>A+2C=2\rightarrow</math> 3 | + | <math>A+2C=2\rightarrow</math> 3 ways |
<math>A+2C=2\rightarrow</math> 2 ways | <math>A+2C=2\rightarrow</math> 2 ways | ||
| − | + | We have a total of <math>18</math> ways. | |
| − | |||
==See also== | ==See also== | ||
{{ARML box|year=2005|state=Alabama|num-b=3|num-a=5}} | {{ARML box|year=2005|state=Alabama|num-b=3|num-a=5}} | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
| + | |||
| + | [[Category:Intermediate Combinatorics Problems]] | ||
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 | ||
