Difference between revisions of "2022 AMC 10B Problems/Problem 5"
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<math>\textbf{(A)}\ \sqrt3 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ \sqrt{15} \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \sqrt{105}</math> | <math>\textbf{(A)}\ \sqrt3 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ \sqrt{15} \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \sqrt{105}</math> | ||
− | ==Solution 1 | + | ==Solution 1== |
We apply the difference of squares to the denominator, and then regroup factors: | We apply the difference of squares to the denominator, and then regroup factors: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} |
Revision as of 12:20, 28 November 2022
Contents
Problem
What is the value of
Solution 1
We apply the difference of squares to the denominator, and then regroup factors: ~MRENTHUSIASM
Solution 2 (Brute Force)
Since these numbers are fairly small, we can use brute force as follows: ~not_slay
Solution 3 (Brute Force)
This solution starts off exactly as the one above. We simplify to get:
But now, we can get a nice simplification as shown:
~TaeKim
~minor edits by mathboy100
Video Solution 1
~Education, the Study of Everything
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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 AMC 10 Problems and Solutions |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 4 |
Followed by Problem 6 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.