Difference between revisions of "2017 AMC 8 Problems/Problem 5"
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| − | ==Problem | + | ==Problem== |
What is the value of the expression <math>\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}</math>? | What is the value of the expression <math>\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}</math>? | ||
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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== | ||
https://youtu.be/cY4NYSAD0vQ | https://youtu.be/cY4NYSAD0vQ | ||
| − | ~ | + | https://youtu.be/tGa0YTskTsw |
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/TkZvMa30Juo?t=3529 | ||
| + | |||
| + | ~pi_is_3.14 | ||
==See Also== | ==See Also== | ||
Latest revision as of 13:40, 2 November 2024
Contents
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
