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

Revision as of 19:40, 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 1

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

Solution 2

$2(81+83+85+87+89+91+93+95+97+99)$ Pair the least with the greatest, second least with the second greatest, etc, until you have five pairs, each adding up to $81+99$ = $83+97$ = $85+95$ = $87+93$ = $89+91$ = $180$ . Since we have $5$ pairs, we multiply $180$ by $5$ to get $900$ . But since we have to multiply by 2 (remember the 2 at the beginning of the parentheses!), we get $1800$ , which is $\text{(E)}$ .

1987 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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Difference between revisions of "1987 AJHSME Problems/Problem 3"

Revision as of 19:40, 23 November 2016

Contents

Problem

Solution 1

Solution 2

See Also

@@ Line 8: / Line 8: @@
 Find that
 <cmath>(81+83+85+87+89+91+93+95+97+99) = 5 \cdot 180</cmath>
-Thus
+Which gives us
 <cmath>\begin{align*}
 (5 \cdot 180) &= 10 \cdot 180\\