Difference between revisions of "1997 PMWC Problems/Problem I8"

(New page: ==Problem== <math>997-996-995+994+993-992+991-990-989+988+989-986+\cdots+7-6-5+4+3-2+1=?</math> ==Solution== {{solution}})
 
(See Also)
 
(One intermediate revision by the same user not shown)
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==Solution==
 
==Solution==
  
{{solution}}
+
Wee look for a pattern:
 +
 
 +
+--++-+--++-+
 +
 
 +
So the pattern is +--++-. We find the value of one round:
 +
 
 +
<math>(n-(n-1)-(n-2)+(n-3)+(n-4)-(n-5))=1</math>
 +
 
 +
So we just need to find the number of rounds.
 +
 
 +
There are 6 terms per round, and the +1 doesn't belong to a round, so 996/6=166
 +
 
 +
<math>1(166)+1=167</math>
 +
 
 +
 
 +
 
 +
==See Also==
 +
 
 +
{{PMWC box|year=1997|num-b=I7|num-a=I9}}
 +
 
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 13:16, 15 January 2008

Problem

$997-996-995+994+993-992+991-990-989+988+989-986+\cdots+7-6-5+4+3-2+1=?$

Solution

Wee look for a pattern:

+--++-+--++-+

So the pattern is +--++-. We find the value of one round:

$(n-(n-1)-(n-2)+(n-3)+(n-4)-(n-5))=1$

So we just need to find the number of rounds.

There are 6 terms per round, and the +1 doesn't belong to a round, so 996/6=166

$1(166)+1=167$


See Also

1997 PMWC (Problems)
Preceded by
Problem I7
Followed by
Problem I9
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10
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