Difference between revisions of "1999 AHSME Problems/Problem 1"
(→Solution) |
(→Solution 4) |
||
Line 26: | Line 26: | ||
=== Solution 4 === | === Solution 4 === | ||
We proceed with addition, 1 -2 + 3 -4.... Once done we find <math> \mathrm{(E)}</math> | We proceed with addition, 1 -2 + 3 -4.... Once done we find <math> \mathrm{(E)}</math> | ||
− | |||
− | |||
== See also == | == See also == |
Revision as of 23:36, 26 May 2019
Contents
[hide]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.