2017 AMC 10B Problems/Problem 1

Revision as of 02:04, 16 February 2017 by Thunderlight8 (talk | contribs) (Solution)

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

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 $\textbf{(B) }$

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
-
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png