Difference between revisions of "1992 AHSME Problems/Problem 4"
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\text{(C) odd if c is even; even if c is odd} \quad\\ | \text{(C) odd if c is even; even if c is odd} \quad\\ | ||
\text{(D) odd if c is odd; even if c is even} \quad\\ | \text{(D) odd if c is odd; even if c is even} \quad\\ | ||
− | \text{(E) odd if c is not a multiple of 3;even if c is a multiple of 3} </math> | + | \text{(E) odd if c is not a multiple of 3; even if c is a multiple of 3} </math> |
== Solution == | == Solution == |
Latest revision as of 21:33, 8 September 2020
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 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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