Difference between revisions of "1957 AHSME Problems/Problem 12"
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== Solution == | == Solution == | ||
| − | By the rules of [[exponents]], <math>10^{-49}=10*10^{-50}</math>, which clearly exceeds <math>2*10^{-50}</math>. <math>10^{-49}-2*10^{-50}=10*10^{-50}-2*10^{-50}=8*10^{-50}</math>, so we choose answer choice <math>\boxed{\textbf{(C)}}</math>. | + | By the rules of [[exponentiation|exponents]], <math>10^{-49}=10*10^{-50}</math>, which clearly exceeds <math>2*10^{-50}</math>. <math>10^{-49}-2*10^{-50}=10*10^{-50}-2*10^{-50}=8*10^{-50}</math>, so we choose answer choice <math>\boxed{\textbf{(C)}}</math>. |
==See Also== | ==See Also== | ||
Latest revision as of 08:15, 25 July 2024
Problem 12
Comparing the numbers 
and 
we may say:
Solution
By the rules of exponents, 
, which clearly exceeds 
. 
, so we choose answer choice 
.
See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
