Difference between revisions of "2017 AMC 10B Problems/Problem 1"
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| − | {{ | + | Mary thought of a positive two-digit number. She multiplied it by <math>3</math> and added <math>11</math>. Then she switched the digits of the result, obtaining a number between <math>71</math> and <math>75</math>, inclusive. What was Mary's number? |
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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> | ||
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| + | ==Solution== | ||
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| + | Just try out the answer choices. Multiplying <math>12</math> by <math>3</math> and then adding <math>11</math> gives you <math>74</math>, which works, so the answer is <math>\textbf{(B) }</math> | ||
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2017|ab=B|before=-|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Revision as of 01:02, 16 February 2017
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
Just try out the answer choices. Multiplying 
by and then adding gives you 
, which works, so the answer is 
See Also
| 2017 AMC 10B (Problems • Answer Key • Resources) | ||
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Followed by Problem 2 | |
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| All AMC 10 Problems and Solutions | ||
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