Difference between revisions of "1993 AJHSME Problems/Problem 1"
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<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> | ||
| − | ==Solution== | + | ==Solution 1== |
A. The ordered pair <math>{-4,-9}</math> has a product of <math>-4\cdot-9=36</math> | A. The ordered pair <math>{-4,-9}</math> has a product of <math>-4\cdot-9=36</math> | ||
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Since C is the only ordered pair which doesn't equal 36, <math>\boxed{\text{(C)}}</math> is the answer. | Since C is the only ordered pair which doesn't equal 36, <math>\boxed{\text{(C)}}</math> is the answer. | ||
| + | ==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> | ||
==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}} | ||
Revision as of 19:55, 23 July 2021
Contents
Problem
Which pair of numbers does NOT have a product equal to 
?
Solution 1
A. The ordered pair 
has a product of 
B. The ordered pair 
has a product of 
C. The ordered pair 
has a product of 
D. The ordered pair 
has a product of 
E. The ordered pair 
has a product of 
Since C is the only ordered pair which doesn't equal 36, 
is the answer.
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
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. 
