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2017 AMC 10B Problems/Problem 1

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 $1+1 = 2,$ (we can ignore numbers such as $39+11=50$) 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 $\boxed{\bold{(B)} 12}$.

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


Solution 4 (Fastest Way)

Let x be the original number. The last digit of $3x+11$ must be $7$ so the last digit of $3x$ must be $6$. The only answer choice that satisfies this is $\boxed{\textbf{(B)}\ 12}$.

Solution 5

Subtract $11$ from the numbers $71$ through $75$. This yields $71-11 = 60$, $72-11 = 61$, $73-11 = 62$, $74-11 = 63$, and $75-11 = 64$. Of these, the only ones divisible by $3$ are $60$ and $63$. Therefore, the only possible values are $71$ and $74$. Switching the digits of each, we get $17$ and $47$. Subtracting $11$ from each, we get the numbers $6$ and $36$. Dividing each by $3$, we get $2$ and $12$. The only two-digit number is $12$, so the answer is $\boxed{\textbf{(B)}\ 12}$.

~TheGoldenRetriever

Video Solution (HOW TO CRITICALLY THINK!!!)

https://youtu.be/EbFw47vZASs

~Education, the Study of Everything


Video Solution

https://youtu.be/PQnA07go4GM

Video Solution

https://youtu.be/zTGuz6EoBWY

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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