Difference between revisions of "2013 AMC 8 Problems/Problem 15"
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<math>\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90</math> | <math>\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90</math> | ||
− | ==Video Solution== | + | ==Video Solution by OmegaLearn== |
https://youtu.be/7an5wU9Q5hk?t=301 | https://youtu.be/7an5wU9Q5hk?t=301 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/ew7QnjAAHcw ~savannahsolver | ||
==Solution== | ==Solution== | ||
Line 13: | Line 18: | ||
===Solution 2: Process of Elimination=== | ===Solution 2: Process of Elimination=== | ||
− | First, we solve for <math>s</math>. As Solution 1 perfectly states, <math>5^3+6^s=1421</math> can be simplified to <math>125+6^s=1421</math> which simplifies further to <math>6^s=1296</math>. Therefore, <math>s=4</math>. The only answer choice that is divisible by <math>4</math> is <math>\boxed{\textbf{(B)}\ 40}</math>. | + | First, we solve for <math>s</math>. As Solution 1 perfectly states, <math>5^3+6^s=1421</math> can be simplified to <math>125+6^s=1421</math> which simplifies further to <math>6^s=1296</math>. Therefore, <math>s=4</math>. We know that you cannot take a root of any of the numbers raised to <math>p</math>, <math>r</math>, or <math>s</math> and get a rational answer, and none of the answer choices are irrational, so that rules out the possibility that <math>p</math>, <math>r</math>, or <math>s</math> is a fraction. The only answer choice that is divisible by <math>4</math> is <math>\boxed{\textbf{(B)}\ 40}</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=14|num-a=16}} | {{AMC8 box|year=2013|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 03:58, 29 December 2022
Contents
Problem
If ,
, and
, what is the product of
,
, and
?
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=301
~ pi_is_3.14
Video Solution 2
https://youtu.be/ew7QnjAAHcw ~savannahsolver
Solution
Solution 1: Solving
First, we're going to solve for . Start with
. Then, change
to
. Subtract
from both sides to get
and see that
is
. Now, solve for
. Since
,
must equal
, so
. Now, solve for
.
can be simplified to
which simplifies further to
. Therefore,
.
equals
which equals
. So, the answer is
.
Solution 2: Process of Elimination
First, we solve for . As Solution 1 perfectly states,
can be simplified to
which simplifies further to
. Therefore,
. We know that you cannot take a root of any of the numbers raised to
,
, or
and get a rational answer, and none of the answer choices are irrational, so that rules out the possibility that
,
, or
is a fraction. The only answer choice that is divisible by
is
.
See Also
2013 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.