2017 AMC 10B Problems/Problem 1


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 $1+1 = 2,$ (we can ignore numbers such as $39+11=50$ that do not satisfy this since the numbers must be between $71$ and $75$) 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:

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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.

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Therefore, the answer must be the reversed steps applied to $74.$ We have the following:

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$74\rightarrow47\rightarrow36\rightarrow12$ \[\] Therefore, our answer is $\boxed{\bold{(B)} 12}$.

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 $\boxed{\textbf{(B)}\ 12}$.

Solution 3

You can just plug in the numbers to see which one works. When you get to $12$, you multiply by $3$ and add $11$ to get $47$. When you reverse the digits of $47$, you get $74$, which is within the given range. Thus, the answer is $\boxed{\textbf{(B)}\ 12}$.

Video Solution



Video Solution by TheBeautyofMath



See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
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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