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>
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\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 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 $\fbox{A}$.

See also

1992 AHSME (ProblemsAnswer KeyResources)
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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