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

Just try out the answer choices. Multiplying $12$ by $3$ and then adding $11$ and reversing the digits gives you $74$ , which works, so the answer is $\boxed{\textbf{(B)}\ 12}$ .

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 among the answer choices, the correct answer would be $\boxed{\textbf{(B)}\ 12}$ .

2017 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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2017 AMC 10B Problems/Problem 1

Contents

Problem

Solution 1

Solution 2

See Also