Difference between revisions of "1993 AJHSME Problems/Problem 1"

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==Problem==
 
Which pair of numbers does NOT have a product equal to <math>36</math>?
 
Which pair of numbers does NOT have a product equal to <math>36</math>?
 
<math> \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\} </math>
 
<math> \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\} </math>

Latest revision as of 11:05, 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 (ProblemsAnswer KeyResources)
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

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