Difference between revisions of "2012 AMC 12B Problems/Problem 11"
(→Solution) |
Username222 (talk | contribs) |
||
| Line 10: | Line 10: | ||
Either <math>B = A + 1</math> or <math>B = A - 1</math>, so either <math>A^2 - 5A - 6, B = A + 1</math> or <math>A^2 - 5A - 2, B = A - 1</math>. The second case has no integer roots, and the first can be re-expressed as <math>(A-6)(A+1) = 0, B = A + 1</math>. Since A must be positive, <math>A = 6, B = 7</math> and <math>A+B = 13</math>; C. | Either <math>B = A + 1</math> or <math>B = A - 1</math>, so either <math>A^2 - 5A - 6, B = A + 1</math> or <math>A^2 - 5A - 2, B = A - 1</math>. The second case has no integer roots, and the first can be re-expressed as <math>(A-6)(A+1) = 0, B = A + 1</math>. Since A must be positive, <math>A = 6, B = 7</math> and <math>A+B = 13</math>; C. | ||
| + | == See Also == | ||
| + | |||
| + | {{AMC12 box|year=2012|ab=B|num-b=10|num-a=12}} | ||
Revision as of 22:20, 12 January 2013
Problem
In the equation, 132 (base A) + 43 (base B) = 69 (base A+B), A and B are consecutive integers. What is A+B?
Solution
Change the equation to base 10: ![\[A^2 + 3A +2 + 4B +3= 6A + 6B + 9\]](http://latex.artofproblemsolving.com/7/f/8/7f8c413a644a9cf4e811fa2f03795ac829834483.png)
![\[A^2 - 3A - 2B - 4=0\]](http://latex.artofproblemsolving.com/6/a/d/6adfef05e4387aa3181bf060ddd0e5b33da04966.png)
Either 
or 
, so either 
or 
. The second case has no integer roots, and the first can be re-expressed as 
. Since A must be positive, 
and 
; C.
See Also
| 2012 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
