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

Revision as of 14:41, 25 November 2017

Problem

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

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

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 $-49 + 99 = 50 \Rightarrow \mathrm{(E)}$ .

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 $1+49 = 50 \Rightarrow \mathrm{(E)}$ .

Solution 3

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

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

We add:

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

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

1999 AHSME (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 9: / Line 9: @@
 === 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>.

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