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Difference between revisions of "2015 AMC 12B Problems/Problem 3"

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==Solution==
 
==Solution==
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Let <math>a</math> be the number written two times, and <math>b</math> the number written three times. Then <math>2a + 3b = 100</math>. Plugging in <math>a</math> = 28 doesn't yield an integer for <math>b</math>, so it must be that <math>b = 28</math>, and we get <math>2a + 84 = 100</math>. Solving for <math>a</math>, we obtain <math>a = \boxed{\textbf{(A)}\; 8}</math>.
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Let <math>a</math> be the number written two times, and <math>b</math> the number written three times. Then <math>2a + 3b = 100</math>. Plugging in <math>a = 28</math> doesn't yield an integer for <math>b</math>, so it must be that <math>b = 28</math>, and we get <math>2a + 84 = 100</math>. Solving for <math>a</math>, we obtain <math>a = \boxed{\textbf{(A)}\; 8}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2015|ab=B|num-a=4|num-b=2}}
 
{{AMC12 box|year=2015|ab=B|num-a=4|num-b=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:33, 20 January 2018

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 (Problems • Answer Key • Resources)
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

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