Difference between revisions of "1985 AJHSME Problems/Problem 3"
(→Solution 2 (Brute Force)) |
|||
(9 intermediate revisions by 5 users not shown) | |||
Line 8: | Line 8: | ||
==Solution== | ==Solution== | ||
− | We immediately see some canceling. We see powers of ten on the top and on the bottom of the fraction, and we | + | We immediately see some canceling. We see powers of ten on the top and on the bottom of the fraction, and we make quick work of this: <cmath>\frac{10^7}{5 \times 10^4} = \frac{10^3}{5}</cmath> |
− | So the answer is (D) | + | We know that <math>10^3 = 10 \times 10 \times 10</math>, so |
+ | |||
+ | <cmath>\begin{align*} | ||
+ | \frac{10^3}{5} &= \frac{10\times 10\times 10}{5} \\ | ||
+ | &= 2\times 10\times 10 \\ | ||
+ | &= 200 \\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | So the answer is <math>\boxed{\text{D}}</math> | ||
+ | |||
+ | == Solution 2 (Brute Force) == | ||
+ | |||
+ | <math>10^7</math> is <math>10000000</math> | ||
+ | |||
+ | <math>5 \times 10^4</math> is <math>50000</math> | ||
+ | |||
+ | Thus the answer is <math>200</math>, or <math>\boxed{\textbf{(D)}\ 200}</math> | ||
+ | |||
+ | ==Video Solution by BoundlessBrain!== | ||
+ | https://youtu.be/BfFv227egOg | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/KW5HexBjEHU | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
− | [[ | + | {{AJHSME box|year=1985|num-b=2|num-a=4}} |
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | |||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 22:07, 2 October 2024
Contents
Problem
Solution
We immediately see some canceling. We see powers of ten on the top and on the bottom of the fraction, and we make quick work of this:
We know that , so
So the answer is
Solution 2 (Brute Force)
is
is
Thus the answer is , or
Video Solution by BoundlessBrain!
Video Solution
~savannahsolver
See Also
1985 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.