Difference between revisions of "1987 AJHSME Problems/Problem 3"
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<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>\text{(E)}</math>. |
==See Also== | ==See Also== | ||
Revision as of 18:18, 23 November 2016
Problem
Solution
Pair the least with the greatest, second least with the second greatest, etc, until you have five pairs, each adding up to 
= 
= 
= 
= 
= 
. Since we have 
pairs, we multiply by to get 
. But since we have to multiply by 2 (remember the 2 at the beginning of the parentheses!), we get 
, which is 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
