2012 AMC 12B Problems/Problem 11: Difference between revisions
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We can eliminate answer choice <math>\textbf{(A)}</math> because you can't have a <math>9</math> in base <math>9</math>. Now we know that A and B are consecutive, so we can just test answers. You will only have to test at most <math>8</math> cases. Eventually, after testing a few cases, you will find that <math>A=6</math> and <math>B=7</math>. The solution is <math>\boxed{\mathbf{(C)}\ 13}</math>. | We can eliminate answer choice <math>\textbf{(A)}</math> because you can't have a <math>9</math> in base <math>9</math>. Now we know that A and B are consecutive, so we can just test answers. You will only have to test at most <math>8</math> cases. Eventually, after testing a few cases, you will find that <math>A=6</math> and <math>B=7</math>. The solution is <math>\boxed{\mathbf{(C)}\ 13}</math>. | ||
~ | ~kempwood | ||
== See Also == | == See Also == | ||
Latest revision as of 20:05, 23 December 2023
Problem
In the equation below, 
and 
are consecutive positive integers, and , , and 
represent number bases: ![\[132_A+43_B=69_{A+B}.\]](http://latex.artofproblemsolving.com/8/b/c/8bc820c4707b39c3a7159da77442c6e2c08e8c3b.png)
What is ?
Solution 1
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 
Solution 2 (Answer Choices)
We can eliminate answer choice 
because you can't have a 
in base . Now we know that A and B are consecutive, so we can just test answers. You will only have to test at most 
cases. Eventually, after testing a few cases, you will find that 
and 
. The solution is 
.
~kempwood
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 | |
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