2015 AMC 12B Problems/Problem 3

Revision as of 22:33, 20 January 2018 by Flyhawkeye (talk | contribs) (Solution: changed latex)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Isaac has written down one integer two times and another integer three times. The sum of the five numbers is 100, and one of the numbers is 28. What is the other number?

$\textbf{(A)}\; 8 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 14 \qquad\textbf{(D)}\; 15 \qquad\textbf{(E)}\; 18$

Solution

Let $a$ be the number written two times, and $b$ the number written three times. Then $2a + 3b = 100$. Plugging in $a = 28$ doesn't yield an integer for $b$, so it must be that $b = 28$, and we get $2a + 84 = 100$. Solving for $a$, we obtain $a = \boxed{\textbf{(A)}\; 8}$.

See Also

2015 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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 12 Problems and Solutions

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