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Difference between revisions of "2017 AMC 10B Problems/Problem 1"
(Adding solution 2.)
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<math>\textbf{(A)}\ 11\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15</math>
 
<math>\textbf{(A)}\ 11\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15</math>
  
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==Solution==
+
==Solutions==
  
 +
===Solution 1===
 
Just try out the answer choices. Multiplying <math>12</math> by <math>3</math> and then adding <math>11</math> and reversing the digits gives you <math>74</math>, which works, so the answer is  <math>\textbf{(B) }</math>
 
Just try out the answer choices. Multiplying <math>12</math> by <math>3</math> and then adding <math>11</math> and reversing the digits gives you <math>74</math>, which works, so the answer is  <math>\textbf{(B) }</math>
 +
 +
===Solution 2===
 +
Working backwards, we reverse the digits of each number from <math>71</math>~<math>75</math> and subtract <math>11</math> from each, so we have
 +
<cmath>6, 16, 26, 36, 46</cmath>
 +
The only numbers from this list that are divisible by <math>3</math> are <math>6</math> and <math>36</math>. We divide both by <math>3</math>, yielding <math>2</math> and <math>12</math>. Since <math>2</math> is not among the answer choices, the correct answer would be <math>\boxed{\textbf{(B)}12.}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|before=-|num-a=2}}
 
{{AMC10 box|year=2017|ab=B|before=-|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 07:49, 16 February 2017

Problem

Mary thought of a positive two-digit number. She multiplied it by $3$ and added $11$. Then she switched the digits of the result, obtaining a number between $71$ and $75$, inclusive. What was Mary's number?

$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$

Solutions

Solution 1

Just try out the answer choices. Multiplying $12$ by $3$ and then adding $11$ and reversing the digits gives you $74$, which works, so the answer is $\textbf{(B) }$

Solution 2

Working backwards, we reverse the digits of each number from $71$~$75$ and subtract $11$ from each, so we have \[6, 16, 26, 36, 46\] The only numbers from this list that are divisible by $3$ are $6$ and $36$. We divide both by $3$, yielding $2$ and $12$. Since $2$ is not among the answer choices, the correct answer would be $\boxed{\textbf{(B)}12.}$

See Also

2017 AMC 10B (Problems • Answer Key • Resources)
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Problem 2
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 10 Problems and Solutions

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