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

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==Solution 2==
 
==Solution 2==
 
<math>2(81+83+85+87+89+91+93+95+97+99)</math>
 
<math>2(81+83+85+87+89+91+93+95+97+99)</math>
−
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>.
+
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>\boxed{\text{E}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 13:30, 15 March 2018

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 \boxed{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 $\boxed{\text{E}}$.

See Also

1987 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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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