Difference between revisions of "2018 AMC 8 Problems/Problem 2"

Revision as of 16:10, 21 November 2018

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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

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-==Problem 2==
+==Problem 24==
-What is the value of the product<cmath>\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?</cmath>
-<math>\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{4}{3}\qquad\textbf{(C) }\frac{7}{2}\qquad\textbf{(D) }7\qquad\textbf{(E) }8</math>
 ==Solution==
-You may simply the expression to get <math>\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6}</math>.
-Using telescoping, many things cancel out. You are left with <math>\frac{7}{1}</math> which is equivelant to <math>7</math>, or <math>\textbf{(D) }</math>
 ==See Also==
-{{AMC8 box|year=2018|num-b=1|num-a=3}}
+{{AMC8 box|year=2018|num-b=23|num-a=25}}
 {{MAA Notice}}