Difference between revisions of "2017 AMC 8 Problems/Problem 5"
Elbertpark (talk | contribs) m |
Elbertpark (talk | contribs) (→Solution 3) |
||
Line 16: | Line 16: | ||
==Solution 3== | ==Solution 3== | ||
− | First, we evaluate <math>1 + 2 + 3 + 4 + 5 + 6 + 7 + 8</math>, to get 36. We notice that 36 is 6 squared, so we can | + | First, we evaluate <math>1 + 2 + 3 + 4 + 5 + 6 + 7 + 8</math>, to get 36. We notice that 36 is 6 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, multiplying <math>4 \cdot 5 \cdot 7 \cdot 8 \cdot</math>. Multiplying these, we get <math>\boxed{\textbf{(B)}\ 1120}</math>. |
~[[User:elbertpark|elbertpark]] | ~[[User:elbertpark|elbertpark]] |
Revision as of 01:25, 6 November 2020
Problem 5
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 36 is 6 squared, so we can factor the denominator like then cancel the 6s out, to get . Now that we have escaped fraction form, multiplying . Multiplying these, we get .
Video Solution
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.