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 "1993 AJHSME Problems/Problem 1"
m
 
(5 intermediate revisions by 3 users not shown)
Line 4: Line 4:
  
 
==Solution 1==
 
==Solution 1==
−
A. The ordered pair <math>{-4,-9}</math> has a product of <math>-4\cdot-9=36</math>
 
  
−
B. The ordered pair <math>{-3,-12}</math> has a product of <math>-3\cdot-12=36</math>
+
Let's calculate each of the answer choices and see which one DOES NOT equal <math>36</math>.
  
−
C. The ordered pair <math>{1/12, -72}</math> has a product of <math>-36</math>
+
<math>A</math> comes out to be <math>-4 \times -9= 36</math>,
  
−
D. The ordered pair <math>{1, 36}</math> has a product of <math>1\cdot36=36</math>
+
<math>B</math> equals <math>-3 \times -12= 36</math>,
  
−
E. The ordered pair <math>{3/2, 24}</math> has a product of <math>3/2\cdot24=36</math>
+
<math>C</math> is <math>\frac{1}{2} \times -72= -36</math>,
  
−
Since C is the only ordered pair which doesn't equal 36, <math>\boxed{\text{(C)}}</math> is the answer.
+
<math>D</math> simplifies to <math>1 \times 36= 36</math>,
 +
 
 +
and <math>E</math> equals <math>\frac{3}{2} \times 24= 3 \times 12= 36</math>.
 +
 
 +
Thus, our answer is <math>\boxed{C}</math>.
  
 
==Solution 2==
 
==Solution 2==
−
We know that if we want a product of 36, both numbers have to be positive or negative. Scanning the number pairs, the only choice with one negative number and one positive number is <math>\boxed{\text{(C)}}</math>  
+
 
 +
In order for a product to be positive (<math>36</math>), the numbers should either be both positive or both negative. Looking at our answer choices, the only option that does not fit this description is <math>\boxed{C}</math>.
 +
 
 
==See Also==
 
==See Also==
 
{{AJHSME box|year=1993|before=First<br />Question|num-a=2}}
 
{{AJHSME box|year=1993|before=First<br />Question|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:05, 27 June 2023

Problem

Which pair of numbers does NOT have a product equal to $36$? $\text{(A)}\ \{-4,-9\}\qquad\text{(B)}\ \{-3,-12\}\qquad\text{(C)}\ \left\{\frac{1}{2},-72\right\}\qquad\text{(D)}\ \{ 1,36\}\qquad\text{(E)}\ \left\{\frac{3}{2},24\right\}$

Solution 1

Let's calculate each of the answer choices and see which one DOES NOT equal $36$.

$A$ comes out to be $-4 \times -9= 36$,

$B$ equals $-3 \times -12= 36$,

$C$ is $\frac{1}{2} \times -72= -36$,

$D$ simplifies to $1 \times 36= 36$,

and $E$ equals $\frac{3}{2} \times 24= 3 \times 12= 36$.

Thus, our answer is $\boxed{C}$.

Solution 2

In order for a product to be positive ($36$), the numbers should either be both positive or both negative. Looking at our answer choices, the only option that does not fit this description is $\boxed{C}$.

See Also

1993 AJHSME (Problems • Answer Key • Resources)
Preceded by
First
Question 		Followed by
Problem 2
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
All AJHSME/AMC 8 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=1993_AJHSME_Problems/Problem_1&oldid=194868"