1996 AJHSME Problems/Problem 16

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

See Also

1996 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AJHSME/AMC 8 Problems and Solutions

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