Difference between revisions of "2017 AMC 10B Problems/Problem 1"
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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> | ||
| − | + | ==Solution 1== | |
| − | |||
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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 | 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> | <cmath>6, 16, 26, 36, 46</cmath> | ||
Revision as of 11:37, 16 February 2017
Contents
Problem
Mary thought of a positive two-digit number. She multiplied it by 
and added 
. Then she switched the digits of the result, obtaining a number between 
and 
, inclusive. What was Mary's number?
Solution 1
Just try out the answer choices. Multiplying 
by and then adding and reversing the digits gives you 
, which works, so the answer is 
Solution 2
Working backwards, we reverse the digits of each number from ~ and subtract from each, so we have
![\[6, 16, 26, 36, 46\]](http://latex.artofproblemsolving.com/f/1/4/f1470bf84cc79a1ad9170b8ff6a076ee889b0b71.png)
The only numbers from this list that are divisible by are 
and 
. We divide both by , yielding 
and . Since is not among the answer choices, the correct answer would be 
See Also
| 2017 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
