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Difference between revisions of "2017 AMC 10B Problems/Problem 1"
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==Solution 1==
 
==Solution 1==
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We try out each answer choice. Multiplying <math>12</math> by <math>3</math>, adding <math>11</math>, and reversing the digits yields <math>74</math>. Therefore the answer is <math>\boxed{\textbf{(B)}\ 12}</math>.
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Let her <math>12</math>-digit number be <math>x</math>. Multiplying by <math>3</math> makes it a multiple of <math>3</math>, meaning that the sum of its digits is divisible by <math>3</math>. Adding on <math>11</math> increases the sum of the digits by <math>1+1 = 2,</math> and reversing the digits keeps the sum of the digits the same; this means that the resulting number must be <math>2</math> more than a multiple of <math>3</math>. There are two such numbers between <math>71</math> and <math>75</math>: <math>71</math> and <math>74.</math> Now that we have narrowed down the choices, we can simply test the answers to see which one will provide a two-digit number when the steps are reversed:
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<cmath></cmath>
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For <math>71,</math> we reverse the digits, resulting in <math>17.</math> Subtracting <math>11</math>, we get <math>6.</math> We can already see that dividing this by <math>3</math> will not be a two-digit number, so <math>71</math> does not meet our requirements.
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<cmath></cmath>
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Therefore, the answer must be the reversed steps applied to <math>74.</math> We have the following:
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<math>74\rightarrow47\rightarrow36\rightarrow12</math>
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Therefore, our answer is <math>\boxed{12.}</math>
  
 
==Solution 2==
 
==Solution 2==

Revision as of 15:11, 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?

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

Solution 1

Let her $12$-digit number be $x$. Multiplying by $3$ makes it a multiple of $3$, meaning that the sum of its digits is divisible by $3$. Adding on $11$ increases the sum of the digits by $1+1 = 2,$ and reversing the digits keeps the sum of the digits the same; this means that the resulting number must be $2$ more than a multiple of $3$. There are two such numbers between $71$ and $75$: $71$ and $74.$ Now that we have narrowed down the choices, we can simply test the answers to see which one will provide a two-digit number when the steps are reversed: \[\] For $71,$ we reverse the digits, resulting in $17.$ Subtracting $11$, we get $6.$ We can already see that dividing this by $3$ will not be a two-digit number, so $71$ does not meet our requirements. \[\] Therefore, the answer must be the reversed steps applied to $74.$ We have the following:

$74\rightarrow47\rightarrow36\rightarrow12$

Therefore, our answer is $\boxed{12.}$

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 a two-digit number, the answer is $\boxed{\textbf{(B)}\ 12}$.

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

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