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Difference between revisions of "2017 AMC 10B Problems/Problem 1"

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<cmath></cmath>
 
<cmath></cmath>
 
Therefore, our answer is <math>\boxed{\bold{(B)} 12.}</math>
 
Therefore, our answer is <math>\boxed{\bold{(B)} 12.}</math>
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==Solution 2==
 
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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
 
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<cmath>6, 16, 26, 36, 46</cmath>
 
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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 a two-digit number, the answer is <math>\boxed{\textbf{(B)}\ 12}</math>.
 
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|before=First Problem|num-a=2}}
 
{{AMC10 box|year=2017|ab=B|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:08, 26 July 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 $2$-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 $\boxed{\bold{(B)} 12.}$

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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