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

(Solution)
(Solution)
Line 8: Line 8:
  
 
<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>(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>\text{(E)</math>.
  
 
==See Also==
 
==See Also==

Revision as of 19:18, 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

$2(81+83+85+87+89+91+93+95+97+99)$ 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 $\text{(E)$ (Error compiling LaTeX. Unknown error_msg).

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

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