2017 AMC 8

Complete problem set with solutions and individual problem pages

Problem 5 Easy

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}?

  • A.

    1020

  • B.

    1120

  • C.

    1220

  • D.

    2240

  • E.

    3360

Answer:B

Solution 1

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}.

 

Solution 2

First, we evaluate 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 to get 36. We notice that 36 is 6 squared, so we can factor the denominator like \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{6 \cdot 6} then cancel the 6s out to get \frac{4 \cdot 5 \cdot 7 \cdot 8}{1}. Now that we have escaped fraction form, we multiply 4 \cdot 5 \cdot 7 \cdot 8. Multiplying these, we get \boxed{\textbf{(B)}\ 1120}.

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