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...") |
Golden phi (talk | contribs) (→Solution 2) |
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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>. | ||
Line 16: | Line 16: | ||
Continuing the pattern, every group has a sum of <math>0</math>, and thus the entire sum is <math>0</math>, giving an answer of <math>\boxed{C}</math>. | Continuing the pattern, every group has a sum of <math>0</math>, and thus the entire sum is <math>0</math>, giving an answer of <math>\boxed{C}</math>. | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ~Golden_Phi | ||
==See Also== | ==See Also== | ||
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* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
+ | {{MAA Notice}} |
Latest revision as of 02:20, 29 March 2024
Contents
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 .
Solution 2
Let any term of the series be . Realize that at every , the sum of the series is 0. For we know so the solution is .
~Golden_Phi
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.