Difference between revisions of "1999 AHSME Problems/Problem 1"
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<math>2S=100-100+100-100\cdots +100=100</math> | <math>2S=100-100+100-100\cdots +100=100</math> | ||
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| + | <math>S=50\Rightarrow \mathrm{(E)}</math> | ||
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| + | === Solution 4 === | ||
| + | We proceed with addition, 1 -2 + 3 -4.... Once done we find <math> \mathrm{(E)}</math> | ||
<math>S=50\Rightarrow \mathrm{(E)}</math> | <math>S=50\Rightarrow \mathrm{(E)}</math> | ||
Revision as of 23:35, 26 May 2019
Contents
Problem
Solution
Solution 1
If we group consecutive terms together, we get 
, and since there are 49 pairs of terms the answer is 
.
Solution 2
( Similar to Solution 1 )
If we rearranged the terms, we get 
then 
, and since there are 49 pairs of terms and the 
in the beginning the answer is 
.
Solution 3
Let 
.
Therefore, 
We add:
Solution 4
We proceed with addition, 1 -2 + 3 -4.... Once done we find 
See also
| 1999 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by First question |
Followed by Problem 2 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
