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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>
  
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==Solution==
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==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>.
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==Solution 2 (Answer Choices)==
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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>.
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~Hithere22702
  
 
== See Also ==
 
== See Also ==

Revision as of 11:35, 25 April 2020

Problem

In the equation below, $A$ and $B$ are consecutive positive integers, and $A$, $B$, and $A+B$ represent number bases: \[132_A+43_B=69_{A+B}.\] What is $A+B$?

$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 17$

Solution 1

Change the equation to base 10: \[A^2 + 3A +2 + 4B +3= 6A + 6B + 9\] \[A^2 - 3A - 2B - 4=0\]

Either $B = A + 1$ or $B = A - 1$, so either $A^2 - 5A - 6, B = A + 1$ or $A^2 - 5A - 2, B = A - 1$. The second case has no integer roots, and the first can be re-expressed as $(A-6)(A+1) = 0, B = A + 1$. Since A must be positive, $A = 6, B = 7$ and $A+B = 13$ $\textrm{ (C) }$.

Solution 2 (Answer Choices)

We can eliminate answer choice $\textbf{(A)}$ because you can't have a $9$ in base $9$. Now we know that A and B are consecutive, so we can just test answers. You will only have to test at most $8$ cases. Eventually, after testing a few cases, you will find that $A=6$ and $B=7$. The solution is $\boxed{\mathbf{(C)}\ 13}$.

~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

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