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>. | + | 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]] |
==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}} | ||
Revision as of 19:42, 30 October 2016
Problem
Suppose m and n are positive odd integers. Which of the following must also be an odd integer?
Solution
Assume WLOG that 
and 
are both 
. Plugging into each of the choices, we get 
and 
. The only odd integer is 
. HI
See Also
| 2005 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
