Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Corrected typo in question and added solution)
 
(One intermediate revision by one other user not shown)
Line 7: Line 7:
 
\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 ==
−
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>.
+
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 ==

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 (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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_4&oldid=133371"