Difference between revisions of "2018 AMC 8 Problems/Problem 2"
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Using telescoping, most of the terms cancel out diagonally. We are left with <math>\frac{7}{1}</math> which is equivalent to <math>7</math>. Thus, the answer would be <math>\boxed{\textbf{(D) }7}</math>. | Using telescoping, most of the terms cancel out diagonally. We are left with <math>\frac{7}{1}</math> which is equivalent to <math>7</math>. Thus, the answer would be <math>\boxed{\textbf{(D) }7}</math>. | ||
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+ | == Video Solution (CRITICAL THINKING!!!)== | ||
+ | https://youtu.be/OFsrMjvR950 | ||
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+ | ~Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== | ||
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~savannahsolver | ~savannahsolver | ||
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+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/TkZvMa30Juo?t=3213 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
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==See also== | ==See also== |
Latest revision as of 20:19, 29 March 2023
Contents
[hide]Problem
What is the value of the product
Solution
By adding up the numbers in each of the parentheses, we get:
.
Using telescoping, most of the terms cancel out diagonally. We are left with which is equivalent to . Thus, the answer would be .
Video Solution (CRITICAL THINKING!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/TkZvMa30Juo?t=3213
~ pi_is_3.14
See also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.