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

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

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

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Doesn't Exist	Followed by Problem 2
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All AMC 10 Problems and Solutions

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