Difference between revisions of "1950 AHSME Problems/Problem 26"
(→Solution 2) |
m (→Solution 3) |
||
(One intermediate revision by one other user not shown) | |||
Line 17: | Line 17: | ||
<cmath>10^b=mn</cmath> | <cmath>10^b=mn</cmath> | ||
<cmath>m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}</cmath> | <cmath>m=\boxed{\mathrm{(E) }\dfrac{10^b}{n}}</cmath> | ||
+ | ~Vndom | ||
+ | |||
+ | ==Solution 3== | ||
+ | More simply, we can just simulate the problem, if we have <math>m = 10</math>, that means the right side must be 1, so the only way we can achieve that with distinct <math>n</math>, is if <math>b = 3</math>, and <math>n = 100</math>. With this we can look through the different answer choices substituting in <math>b</math>, <math>n</math>, and <math>m</math>, and find that <math>\boxed{\mathrm{(E)}}</math> is the only one that satisfies the question. | ||
+ | |||
+ | ~Shadow-18 | ||
==See Also== | ==See Also== |
Latest revision as of 18:38, 29 October 2020
Contents
[hide]Problem
If , then
Solution 1
We have . Substituting, we find . Using , the left side becomes . Because , .
Solution 2
adding to both sides: using the logarithm property: rewriting in exponential notation: ~Vndom
Solution 3
More simply, we can just simulate the problem, if we have , that means the right side must be 1, so the only way we can achieve that with distinct , is if , and . With this we can look through the different answer choices substituting in , , and , and find that is the only one that satisfies the question.
~Shadow-18
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.