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> | ||
− | == | + | ==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
Contents
[hide]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?
Solutions
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 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 - |
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 |
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