Difference between revisions of "2005 AMC 8 Problems/Problem 8"

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==Solution==
 
==Solution==
Assume [[WLOG]] that <math>m</math> and <math>n</math> are both <math>1</math>. Plugging into each of the choices, we get <math>4, 2, 6, 16,</math> and <math>3</math>. The only odd integer is <math>\boxed{\textbf{(E)}\ 3mn}</math>. [[HI]]
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Assume [[WLOG]] that <math>m</math> and <math>n</math> are both <math>1</math>. Plugging into each of the choices, we get <math>4, 2, 6, 16,</math> and <math>3</math>. The only odd integer is <math>\boxed{\textbf{(E)}\ 3mn}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2005|num-b=7|num-a=9}}
 
{{AMC8 box|year=2005|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:18, 1 November 2016

Problem

Suppose m and n are positive odd integers. Which of the following must also be an odd integer?

$\textbf{(A)}\ m+3n\qquad\textbf{(B)}\ 3m-n\qquad\textbf{(C)}\ 3m^2 + 3n^2\qquad\textbf{(D)}\ (nm + 3)^2\qquad\textbf{(E)}\ 3mn$

Solution

Assume WLOG that $m$ and $n$ are both $1$. Plugging into each of the choices, we get $4, 2, 6, 16,$ and $3$. The only odd integer is $\boxed{\textbf{(E)}\ 3mn}$.

See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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 AJHSME/AMC 8 Problems and Solutions

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