Difference between revisions of "1983 AHSME Problems/Problem 25"
Quantummech (talk | contribs) (→Solution) |
Sevenoptimus (talk | contribs) m (Cleaned up the solution) |
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==Solution== | ==Solution== | ||
− | We have that <math>12=\frac{60}{5}</math>. We can substitute our value for 5, to get | + | We have that <math>12=\frac{60}{5}</math>. We can substitute our value for 5, to get |
<cmath>12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{(1-b)}.</cmath> | <cmath>12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{(1-b)}.</cmath> | ||
− | Therefore we have | + | Therefore we have |
<cmath>12^{[(1-a-b)/2(1-b)]}=60^{[(1-b)(1-a-b)/2(1-b)]}=60^{(1-a-b)/2}.</cmath> | <cmath>12^{[(1-a-b)/2(1-b)]}=60^{[(1-b)(1-a-b)/2(1-b)]}=60^{(1-a-b)/2}.</cmath> | ||
− | Since <math>4=\frac{60}{3\cdot 5}</math>, we have | + | Since <math>4=\frac{60}{3\cdot 5}</math>, we have |
<cmath>4=\frac{60}{60^a 60^b}=60^{(1-a-b)}.</cmath> | <cmath>4=\frac{60}{60^a 60^b}=60^{(1-a-b)}.</cmath> | ||
− | Therefore, we have | + | Therefore, we have |
<cmath>60^{(1-a-b)/2}=4^{1/2}=\fbox{\textbf{(B)} 2}</cmath> | <cmath>60^{(1-a-b)/2}=4^{1/2}=\fbox{\textbf{(B)} 2}</cmath> | ||
Revision as of 18:25, 27 January 2019
Problem 25
If and
, then
is
Solution
We have that . We can substitute our value for 5, to get
Therefore we have
Since
, we have
Therefore, we have
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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.