1999 AHSME Problems/Problem 1: Difference between revisions
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<math>1 - 2 + 3 -4 + \cdots - 98 + 99 = </math> | <math>1 - 2 + 3 -4 + \cdots - 98 + 99 = </math> | ||
<math> \mathrm{(A) \ -50 } \qquad \mathrm{(B) \ -49 } \qquad \mathrm{(C) \ 0 } \qquad \mathrm{(D) \ 49 } \qquad \mathrm{(E) \ 50 } </math> | <math> \mathrm{\textbf{(A)} \ -50 } \qquad \mathrm{\textbf{(B)} \ -49 } \qquad \mathrm{\textbf{(C)} \ 0 } \qquad \mathrm{\textbf{(D)} \ 49 } \qquad \mathrm{\textbf{(E)} \ 50 } </math> | ||
== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
If we group consecutive terms together, we get <math>(-1) + (-1) + \cdots + 99</math>, and since there are 49 pairs of terms the answer is <math>-49 + 99 = 50 \Rightarrow \mathrm{(E)}</math>. | If we group consecutive terms together, we get <math>(-1) + (-1) + \cdots + 99</math>, and since there are 49 pairs of terms the answer is <math>-49 + 99 = 50 \Rightarrow \mathrm{\textbf{(E)}}</math>. | ||
=== 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{\textbf(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>. | ||
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<math>2S=100-100+100-100\cdots +100=100</math> | <math>2S=100-100+100-100\cdots +100=100</math> | ||
<math>S=50\Rightarrow \mathrm{(E)}</math> | <math>S=50\Rightarrow \mathrm{\textbf{(E)}}</math> | ||
=== Solution 4 === | |||
We proceed with addition, 1 -2 + 3 -4.... Once done we find <math> \mathrm{\textbf{(E)}}</math> | |||
== See also == | == See also == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | |||
Latest revision as of 20:35, 9 February 2024
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
