Difference between revisions of "1996 AJHSME Problems/Problem 16"

Latest revision as of 21:35, 27 November 2024

Problem

$1-2-3+4+5-6-7+8+9-10-11+\cdots + 1992+1993-1994-1995+1996=$

$\text{(A)}\ -998 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 998$

Solution

Put the numbers in groups of $4$ :

$(1-2-3+4)+(5-6-7+8)+(9-10-11+ 12) + \cdots + (1993-1994-1995+1996)$

The first group has a sum of $0$ .

The second group increases the two positive numbers on the end by $1$ , and decreases the two negative numbers in the middle by $1$ . Thus, the second group also has a sum of $0$ .

Continuing the pattern, every group has a sum of $0$ , and thus the entire sum is $0$ , giving an answer of $\boxed{C}$ .

Solution 2

Let any term of the series be $t_n$ . Realize that at every $n\equiv0 \pmod4$ , the sum of the series is 0. For $t_{1996}$ we know $1996\equiv0 \pmod4$ so the solution is $\boxed{C}$ .

~Golden_Phi

Solution 3

This solution is slightly more detailed than Solution 2, even though they are essentially the same.

Compute the sum of the first and last term, the second and second last term, the third and third last term. We get $(1+1996)+(-2-1995)+(-3-1994)+(4+1993)$ and so on. We see that each pair's sum alternates between $1997$ and $-1997$ .We see that every two consecutive pair cancels out. For example $(1+1996)+(-2-1995)$ gives a result of $0$ . There are a total of $1996$ terms, there are $998$ pairs, and there are $499$ pairs which result in a sum of $1997$ each and $499$ pairs which result in a sum of $-1997$ each. These all cancel out to a sum of $0$ , and thus the entire sum is $0$ , giving an answer of $\boxed{C}$ .

~blankbox

1996 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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
All AJHSME/AMC 8 Problems and Solutions

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

@@ Line 19: / Line 19: @@
 ==Solution 2==
-Let any term of the series be <math>t_n</math>. Realize that at every <math>n\equiv0 \pmod4</math>, the sum of the series is 0. For <math>t_1996</math> we know <math>1996\equiv0 \pmod4</math> so the solution is <math>\boxed{C}</math>.
+Let any term of the series be <math>t_n</math>. Realize that at every <math>n\equiv0 \pmod4</math>, the sum of the series is 0. For <math>t_{1996}</math> we know <math>1996\equiv0 \pmod4</math> so the solution is <math>\boxed{C}</math>.
 ~Golden_Phi
+==Solution 3==
+This solution is slightly more detailed than Solution 2, even though they are essentially the same.
+Compute the sum of the first and last term, the second and second last term, the third and third last term. We get <math>(1+1996)+(-2-1995)+(-3-1994)+(4+1993)</math> and so on. We see that each pair's sum alternates between <math>1997</math> and <math>-1997</math>.We see that every two consecutive pair cancels out. For example <math>(1+1996)+(-2-1995)</math> gives a result of <math>0</math>. There are a total of <math>1996</math> terms, there are <math>998</math> pairs, and there are <math>499</math> pairs which result in a sum of <math>1997</math> each and <math>499</math> pairs which result in a sum of <math>-1997</math> each. These all cancel out to a sum of <math>0</math>, and thus the entire sum is <math>0</math>, giving an answer of <math>\boxed{C}</math>.
+~blankbox
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