Difference between revisions of "2005 Alabama ARML TST Problems/Problem 4"
m |
|||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | For how many ordered | + | For how many [[ordered pair]]s of [[digit]]s <math>\displaystyle (A,B)</math> is <math>\displaystyle 2AB8</math> a [[multiple]] of 12? |
==Solution== | ==Solution== | ||
| − | We wish for <math>2000+100A+10B+8 \equiv 0 \pmod 12\ | + | 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: |
<math>A+2C=2\rightarrow</math> 2 ways | <math>A+2C=2\rightarrow</math> 2 ways | ||
| Line 18: | Line 18: | ||
Total of 18 ways. | Total of 18 ways. | ||
| + | |||
| + | {{wikify}} | ||
==See also== | ==See also== | ||
*[[2005 Alabama ARML TST]] | *[[2005 Alabama ARML TST]] | ||
*[[2005 Alabama ARML TST/Problem 3 | Previous Problem]] | *[[2005 Alabama ARML TST/Problem 3 | Previous Problem]] | ||
*[[2005 Alabama ARML TST/Problem 5 | Next Problem]] | *[[2005 Alabama ARML TST/Problem 5 | Next Problem]] | ||
Revision as of 18:31, 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.
