Difference between revisions of "1983 AHSME Problems/Problem 25"
Sevenoptimus (talk | contribs) m (Fixed problem statement to match the problems page) |
Sevenoptimus (talk | contribs) m (Further cleaned up the solution) |
||
Line 8: | Line 8: | ||
<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^{ | + | <cmath>12^{(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-b)(1-a-b)/\left(2\left(1-b\right)\right)}=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> |
Revision as of 18:28, 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.