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Difference between revisions of "1999 AHSME Problems/Problem 1"

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<math>1 - 2 + 3 -4 + \cdots - 98 + 99 = </math>
 
<math>1 - 2 + 3 -4 + \cdots - 98 + 99 = </math>
  
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<math> \mathrm{(A) \ -50 } \qquad \mathrm{(B) \ -49 } \qquad \mathrm{(C) \ 0 } \qquad \mathrm{(D) \ 49 } \qquad \mathrm{(E) \ 50 }  </math>
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<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 ==
−
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>.
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=== 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{\textbf{(E)}}</math>.
 +
 
 +
=== 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>.
 +
 
 +
Therefore, <math>S=99-98+97-\cdots -4+3-2+1</math>
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 +
We add:
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 +
<math>2S=100-100+100-100\cdots +100=100</math>
 +
 
 +
<math>S=50\Rightarrow \mathrm{\textbf{(E)}}</math>
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 +
 
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=== Solution 4 ===
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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]]
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{{MAA Notice}}

Latest revision as of 20:35, 9 February 2024

Problem

$1 - 2 + 3 -4 + \cdots - 98 + 99 =$

$\mathrm{\textbf{(A)} \ -50 } \qquad \mathrm{\textbf{(B)} \ -49 } \qquad \mathrm{\textbf{(C)} \ 0 } \qquad \mathrm{\textbf{(D)} \ 49 } \qquad \mathrm{\textbf{(E)} \ 50 }$

Solution

Solution 1

If we group consecutive terms together, we get $(-1) + (-1) + \cdots + 99$, and since there are 49 pairs of terms the answer is $-49 + 99 = 50 \Rightarrow \mathrm{\textbf{(E)}}$.

Solution 2

( Similar to Solution 1 ) If we rearranged the terms, we get $1+3-2+5-4 \cdots + 99-98$ then $1 + 1 + \cdots + 1$, and since there are 49 pairs of terms and the $1$ in the beginning the answer is $1+49 = 50 \Rightarrow \mathrm{\textbf(E)}$.

Solution 3

Let $1 - 2 + 3 -4 + \cdots - 98 + 99 = S$.

Therefore, $S=99-98+97-\cdots -4+3-2+1$

We add:

$2S=100-100+100-100\cdots +100=100$

$S=50\Rightarrow \mathrm{\textbf{(E)}}$


Solution 4

We proceed with addition, 1 -2 + 3 -4.... Once done we find $\mathrm{\textbf{(E)}}$

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

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