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Sum the integers, get squares? (Posted on 2007-04-05) Difficulty: 2 of 5
Can the sum of any 2112 consecutive positive integers be a perfect square?

  Submitted by K Sengupta    
Rating: 2.0000 (3 votes)
Solution: (Hide)
(p+1) + (p+2) + ... + (p+2112)
= (p+2112)(p+2113)/2 - p(p+1)/2
= 2112p + 2112.2113/2
= 1056(2p + 2113).

But 1056 = 32.33 and (2p+2113) is odd, so the highest power of 2 dividing 1056(2p+2113) is 32 = 2^5.

Hence the sum of 2112 consecutive positive integers cannot be a perfect square.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSolutionPraneeth2007-09-12 08:19:53
Some Thoughtsi like itAdy TZIDON2007-04-05 14:27:59
SolutionsolutionCharlie2007-04-05 10:25:45
Different methodMarc2007-04-05 10:09:15
SolutionNo SubjectAdy TZIDON2007-04-05 09:24:16
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