2017 AMC 8 Problems/Problem 5: Difference between revisions
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==Solution 2== | ==Solution 2== | ||
It is well known that the sum of all numbers from <math>1</math> to <math>n</math> is <math>\frac{n(n+1)}{2}</math>. Therefore, the denominator is equal to <math>\frac{8 \cdot 9}{2} = 4 \cdot 9 = 2 \cdot 3 \cdot 6</math>. Now we can cancel the factors of <math>2</math>, <math>3</math>, and <math>6</math> from both the numerator and denominator, only leaving <math>8 \cdot 7 \cdot 5 \cdot 4 \cdot 1</math>. This evaluates to <math>\boxed{\textbf{(B)}\ 1120}</math>. | It is well known that the sum of all numbers from <math>1</math> to <math>n</math> is <math>\frac{n(n+1)}{2}</math>. Therefore, the denominator is equal to <math>\frac{8 \cdot 9}{2} = 4 \cdot 9 = 2 \cdot 3 \cdot 6</math>. Now, we can cancel the factors of <math>2</math>, <math>3</math>, and <math>6</math> from both the numerator and denominator, only leaving <math>8 \cdot 7 \cdot 5 \cdot 4 \cdot 1</math>. This evaluates to <math>\boxed{\textbf{(B)}\ 1120}</math>. | ||
==Solution 3== | ==Solution 3== | ||
First, we evaluate <math>1 + 2 + 3 + 4 + 5 + 6 + 7 + 8</math> | First, we evaluate <math>1 + 2 + 3 + 4 + 5 + 6 + 7 + 8</math> to get 36. We notice that <math>36</math> is <math>6</math> squared, so we can factor the denominator like <math>\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{6 \cdot 6}</math> then cancel the 6s out to get <math>\frac{4 \cdot 5 \cdot 7 \cdot 8}{1}</math>. Now that we have escaped fraction form, we multiply <math>4 \cdot 5 \cdot 7 \cdot 8</math>. Multiplying these, we get <math>\boxed{\textbf{(B)}\ 1120}</math>. | ||
~megaboy6679 for minor edits | |||
==Video Solution (CREATIVE THINKING!!!)== | |||
https://youtu.be/O5thBbTXpeY | |||
~Education, the Study of Everything | |||
==Video Solution== | ==Video Solution== | ||
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~savannahsolver | ~savannahsolver | ||
==Video Solution by OmegaLearn== | |||
https://youtu.be/TkZvMa30Juo?t=3529 | |||
~pi_is_3.14 | |||
==See Also== | ==See Also== | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
[[Category:Introductory Algebra Problems]] | |||
Latest revision as of 18:09, 7 June 2025
Problem
What is the value of the expression 
?
Solution 1
Directly calculating:
We evaluate both the top and bottom: 
. This simplifies to 
.
Solution 2
It is well known that the sum of all numbers from 
to 
is 
. Therefore, the denominator is equal to 
. Now, we can cancel the factors of 
, 
, and 
from both the numerator and denominator, only leaving 
. This evaluates to .
Solution 3
First, we evaluate 
to get 36. We notice that 
is squared, so we can factor the denominator like 
then cancel the 6s out to get 
. Now that we have escaped fraction form, we multiply 
. Multiplying these, we get .
~megaboy6679 for minor edits
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/TkZvMa30Juo?t=3529
~pi_is_3.14
See Also
| 2017 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
