Difference between revisions of "1994 AHSME Problems/Problem 1"
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Note that <math>a^x\times a^y=a^{x+y}</math>. So <math>4^4\cdot 4^9=4^{13}</math> and <math>9^4\cdot 9^9=9^{13}</math>. Therefore, <math>4^{13}\cdot 9^{13}=(4\cdot 9)^{13}=\boxed{\textbf{(C)}\ 36^{13}}</math>. | Note that <math>a^x\times a^y=a^{x+y}</math>. So <math>4^4\cdot 4^9=4^{13}</math> and <math>9^4\cdot 9^9=9^{13}</math>. Therefore, <math>4^{13}\cdot 9^{13}=(4\cdot 9)^{13}=\boxed{\textbf{(C)}\ 36^{13}}</math>. | ||
| − | --Solution by [ | + | --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] |
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AHSME box|year=1994|before=First Problem|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:25, 9 January 2021
Problem
Solution
Note that 
. So 
and 
. Therefore, 
.
--Solution by TheMaskedMagician
See Also
| 1994 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
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 • 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. 
