Difference between revisions of "1999 AHSME Problems/Problem 1"
(→Solution 4) |
|||
(3 intermediate revisions by 3 users not shown) | |||
Line 9: | Line 9: | ||
=== Solution 2 === | === Solution 2 === | ||
+ | ( Similar to Solution 1 ) | ||
+ | If we rearranged the terms, we get <math>1+3-2+5-4 \cdots + 99-98</math> then <math>1 + 1 + \cdots + 1 </math>, and since there are 49 pairs of terms and the <math>1</math> in the beginning the answer is <math>1+49 = 50 \Rightarrow \mathrm{(E)}</math>. | ||
+ | |||
+ | === Solution 3 === | ||
Let <math>1 - 2 + 3 -4 + \cdots - 98 + 99 = S</math>. | Let <math>1 - 2 + 3 -4 + \cdots - 98 + 99 = S</math>. | ||
Line 18: | Line 22: | ||
<math>S=50\Rightarrow \mathrm{(E)}</math> | <math>S=50\Rightarrow \mathrm{(E)}</math> | ||
+ | |||
+ | |||
+ | === Solution 4 === | ||
+ | We proceed with addition, 1 -2 + 3 -4.... Once done we find <math> \mathrm{(E)}</math> | ||
== See also == | == See also == | ||
Line 23: | Line 31: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 00:36, 27 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.