2017 AMC 8 Problems/Problem 5: Difference between revisions

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Problem 5

What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$ ?

$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$

Solution 1

Directly calculating:

We evaluate both the top and bottom: $\frac{40320}{36}$ . This simplifies to $\boxed{\textbf{(B)}\ 1120}$ .

Solution 2

It is well known that the sum of all numbers from $1$ to $n$ is $\frac{n(n+1)}{2}$ . Therefore, the denominator is equal to $\frac{8 \cdot 9}{2} = 4 \cdot 9 = 2 \cdot 3 \cdot 6$ . Now we can cancel the factors of $2$ , $3$ , and $6$ from both the numerator and denominator, only leaving $8 \cdot 7 \cdot 5 \cdot 4 \cdot 1$ . This evaluates to $\boxed{\textbf{(B)}\ 1120}$ .

Video Solution

https://youtu.be/cY4NYSAD0vQ

~icematrix2

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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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: File missing

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 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>.
+==Video Solution==
+https://youtu.be/cY4NYSAD0vQ
+~[[User:icematrix2|icematrix2]]
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