Difference between revisions of "1992 AHSME Problems/Problem 4"
(Created page with "== Problem == If <math>a,b</math> and <math>c</math> are positive integers and <math>a</math> and <math>b</math> are odd, then <math>3^a+(b-1)^2c</math> is <math>\text{(A) odd ...") |
(Corrected typo in question and added solution) |
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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; | + | \text{(E) odd if c is not a multiple of 3;even if c is a multiple of 3} </math> |
== Solution == | == Solution == | ||
− | <math>\fbox{A}</math> | + | Since 3 has no factors of 2, <math>3^a</math> will be odd for all values of <math>a</math>. Since <math>b</math> is odd as well, <math>b-1</math> must be even, so <math>(b-1)^2</math> must be even. This means that for all choices of <math>c</math>, <math>(b-1)^2c</math> must be even because any integer times an even number is still even. Since an odd number minus an even number is odd, <math>3^a+(b-1)^2c</math> must be odd for all choices of <math>c</math>, which corresponds to answer choice <math>\fbox{A}</math>. |
== See also == | == See also == |
Revision as of 22:45, 4 October 2016
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 minus 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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All AHSME Problems and Solutions |
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