Difference between revisions of "2017 AMC 8 Problems/Problem 5"
(Created page with "==Problem 5== 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>? <math>\textbf{(A) }1020\qqua...") |
Hashtagmath (talk | contribs) |
||
(16 intermediate revisions by 8 users not shown) | |||
Line 1: | Line 1: | ||
− | ==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>? | ||
<math>\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360</math> | <math>\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | We evaluate both the top and bottom: <math>\frac{40320}{36}</math>. This simplifies to <math>1120</math>, | + | |
+ | Directly calculating: | ||
+ | |||
+ | We evaluate both the top and bottom: <math>\frac{40320}{36}</math>. This simplifies to <math>\boxed{\textbf{(B)}\ 1120}</math>. | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ==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 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]] | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/cY4NYSAD0vQ | ||
==See Also== | ==See Also== | ||
− | {{AMC8 box|year=2017|num-b= | + | {{AMC8 box|year=2017|num-b=4|num-a=6}} |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:59, 18 January 2021
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 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.