1992 AHSME Problems/Problem 4
Problem
If and are positive integers and and are odd, then is
Solution
Since 3 has no factors of 2, will be odd for all values of . Since is odd as well, must be even, so must be even. This means that for all choices of , must be even because any integer times an even number is still even. Since an odd number plus an even number is odd, must be odd for all choices of , which corresponds to answer choice .
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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