Difference between revisions of "1996 AJHSME Problems/Problem 16"
Talkinaway (talk | contribs) (Created page with "==Problem== <math>1-2-3+4+5-6-7+8+9-10-11+\cdots + 1992+1993-1994-1995+1996=</math> <math>\text{(A)}\ -998 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qqua...") |
Talkinaway (talk | contribs) (→Solution) |
||
| Line 9: | Line 9: | ||
Put the numbers in groups of <math>4</math>: | Put the numbers in groups of <math>4</math>: | ||
| − | <math>(1-2-3+4)+(5-6-7+8)+(9-10-11+ 12) + \cdots + (1993-1994-1995+1996) | + | <math>(1-2-3+4)+(5-6-7+8)+(9-10-11+ 12) + \cdots + (1993-1994-1995+1996)</math> |
The first group has a sum of <math>0</math>. | The first group has a sum of <math>0</math>. | ||
Revision as of 17:19, 1 August 2011
Problem
Solution
Put the numbers in groups of 
:
The first group has a sum of 
.
The second group increases the two positive numbers on the end by 
, and decreases the two negative numbers in the middle by . Thus, the second group also has a sum of .
Continuing the pattern, every group has a sum of , and thus the entire sum is , giving an answer of 
.
See Also
| 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 | ||
