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

(Solution 2)
(Solution 2)
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==Solution 2==
 
==Solution 2==
 
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Find that
<math>2(81+83+85+87+89+91+93+95+97+99)</math>
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<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>.
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Thus
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<cmath>\begin{align*}
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2(5 \cdot 180) &= 10 \cdot 180\\
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&= 1800 & \text{ Thus (E) is the correct answer}
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\end{align*}</cmath>
  
 
==Solution 1==
 
==Solution 1==

Revision as of 19:38, 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*}

See Also

1987 AJHSME (ProblemsAnswer KeyResources)
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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