Difference between revisions of "2012 AMC 12B Problems/Problem 11"
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<math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 17 </math> | <math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 17 </math> | ||
− | ==Solution== | + | ==Solution 1== |
Change the equation to base 10: <cmath>A^2 + 3A +2 + 4B +3= 6A + 6B + 9</cmath> <cmath> A^2 - 3A - 2B - 4=0</cmath> | Change the equation to base 10: <cmath>A^2 + 3A +2 + 4B +3= 6A + 6B + 9</cmath> <cmath> A^2 - 3A - 2B - 4=0</cmath> | ||
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> <math> \textrm{ (C) } </math>. | 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> <math> \textrm{ (C) } </math>. | ||
+ | |||
+ | ==Solution 2 (Answer Choices)== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~Hithere22702 | ||
== See Also == | == See Also == |
Revision as of 12:35, 25 April 2020
Problem
In the equation below, and
are consecutive positive integers, and
,
, and
represent number bases:
What is
?
Solution 1
Change the equation to base 10:
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
.
~Hithere22702
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.