Difference between revisions of "1992 AHSME Problems/Problem 4"
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== Solution == | == Solution == | ||
| − | 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 | + | 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 plus 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 02:22, 20 February 2018
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
