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

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<math>\text{(A)}\ 1600 \qquad \text{(B)}\ 1650 \qquad \text{(C)}\ 1700 \qquad \text{(D)}\ 1750 \qquad \text{(E)}\ 1800</math>
 
<math>\text{(A)}\ 1600 \qquad \text{(B)}\ 1650 \qquad \text{(C)}\ 1700 \qquad \text{(D)}\ 1750 \qquad \text{(E)}\ 1800</math>
  
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==Solution==
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==Solution 1==
 +
Find that
 +
<cmath>(81+83+85+87+89+91+93+95+97+99) = 5 \cdot 180</cmath>
 +
Which gives us
 +
<cmath>\begin{align*}
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2(5 \cdot 180) &= 10 \cdot 180\\
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&= 1800 & \text{ Thus \boxed{\text{E}} is the correct answer}
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\end{align*}</cmath>
  
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==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>
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+
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>.
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<math>\begin{align*}
 
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& = 2\Big( (81+99) + (83+97) + (85+95) + (87+93) + (89+91) \Big) \\
 
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& = 2(180+180+180+180+180) \\
 
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& = 2\cdot 5\cdot 180 \\
 
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& = 1800\rightarrow \boxed{\text{E}}
 
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\end{align*}</math>
 
  
 
==See Also==
 
==See Also==
 
{{AJHSME box|year=1987|num-b=2|num-a=4}}
 
{{AJHSME box|year=1987|num-b=2|num-a=4}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 00:03, 22 January 2020

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{\text{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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