Difference between revisions of "1992 AHSME Problems/Problem 4"

Latest revision as of 20:33, 8 September 2020

Problem

If $a,b$ and $c$ are positive integers and $a$ and $b$ are odd, then $3^a+(b-1)^2c$ is

$\text{(A) odd for all choices of c} \quad \text{(B) even for all choices of c} \quad\\ \text{(C) odd if c is even; even if c is odd} \quad\\ \text{(D) odd if c is odd; even if c is even} \quad\\ \text{(E) odd if c is not a multiple of 3; even if c is a multiple of 3}$

Solution

Since 3 has no factors of 2, $3^a$ will be odd for all values of $a$ . Since $b$ is odd as well, $b-1$ must be even, so $(b-1)^2$ must be even. This means that for all choices of $c$ , $(b-1)^2c$ must be even because any integer times an even number is still even. Since an odd number plus an even number is odd, $3^a+(b-1)^2c$ must be odd for all choices of $c$ , which corresponds to answer choice $\fbox{A}$ .

1992 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 7: / Line 7: @@
 \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{(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>
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Difference between revisions of "1992 AHSME Problems/Problem 4"

Latest revision as of 20:33, 8 September 2020

Problem

Solution

See also