Difference between revisions of "2017 AMC 8 Problems/Problem 5"
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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>. | 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>. | ||
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+ | ==Solution 3== | ||
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+ | First, we evaluate 1+2+3+4+5+6+7+8, to get 36. We notice that 36 is 6 squared, so we can go like <math>\frac{\cdot 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{\cdot 4 \cdot 5 \cdot 7 \cdot 8}{1}</math>. Now that we have escaped fraction form, multiplying \cdot 4 \cdot 5 \cdot 7 \cdot 8. Multiplying these, we get <math>\boxed{\textbf{(B)}\ 1120}</math>. | ||
==Video Solution== | ==Video Solution== |
Revision as of 01:19, 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 1+2+3+4+5+6+7+8, to get 36. We notice that 36 is 6 squared, so we can go like then cancel the 6s out, to get . Now that we have escaped fraction form, multiplying \cdot 4 \cdot 5 \cdot 7 \cdot 8. 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.