Difference between revisions of "1987 AJHSME Problems/Problem 3"

Revision as of 20:38, 23 November 2016

Problem

$2(81+83+85+87+89+91+93+95+97+99)=$

$\text{(A)}\ 1600 \qquad \text{(B)}\ 1650 \qquad \text{(C)}\ 1700 \qquad \text{(D)}\ 1750 \qquad \text{(E)}\ 1800$

Solution 2

Find that $(81+83+85+87+89+91+93+95+97+99) = 5 \cdot 180$ Thus $\begin{align*} 2(5 \cdot 180) &= 10 \cdot 180\\ &= 1800 & \text{ Thus (E) is the correct answer} \end{align*}$

Solution 1

Find that $(81+83+85+87+89+91+93+95+97+99) = 5 \cdot 180$ Thus $\begin{align*} 2(5 \cdot 180) &= 10 \cdot 180\\ &= 1800 & \text{ Thus (E) is the correct answer} \end{align*}$

1987 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 6: / Line 6: @@
 ==Solution 2==
+Find that
-<math>2(81+83+85+87+89+91+93+95+97+99)</math>
+<cmath>(81+83+85+87+89+91+93+95+97+99) = 5 \cdot 180</cmath>
-Pair the least with the greatest, second least with the second greatest, etc, until you have five pairs, each adding up to <math>81+99</math> = <math>83+97</math> = <math>85+95</math> = <math>87+93</math> = <math>89+91</math> = <math>180</math>.  Since we have <math>5</math> pairs, we multiply <math>180</math> by <math>5</math> to get <math>900</math>.  But since we have to multiply by 2 (remember the 2 at the beginning of the parentheses!), we get <math>1800</math>, which is <math>\text{(E)}</math>.
+Thus
+<cmath>\begin{align*}
+(5 \cdot 180) &= 10 \cdot 180\\
+&= 1800 & \text{ Thus (E) is the correct answer}
+\end{align*}</cmath>
 ==Solution 1==

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Difference between revisions of "1987 AJHSME Problems/Problem 3"

Revision as of 20:38, 23 November 2016

Contents

Problem

Solution 2

Solution 1

See Also