answersLogoWhite

0

The sum of n consecutive integers is divisible by n when n is odd. It is not divisible by n when n is even. So in this case the answer is it is divisible by 25! Proof: Case I - n is odd: We can substitute 2m+1 (where m is an integer) for n. This lets us produce absolutely any odd integer. Let's look at the sum of any 2m+1 consecutive integers. a + a+d + a+2d + a+3d + ... + a+(n-1)d = n(first+last)/2 (In our problem, the common difference is 1 and this is an arithmetic series.) a + (a+1) + (a+2) + ... + (a+2m) = (2m+1)(2a+2m)/2 = (2m+1)(a+m) It is obvious that this is divisible by 2m+1, our original odd number. That proves case I when n is odd, not for case when it is even. Case II - n is even: We can substitute 2m for n. We have another arithmetic series: a + (a+1) + (a+2) + ... + (a+2m-1) = (2m)(2a+2m-1)/2 = m(2a+2m-1) It is not too hard to prove that this is not divisible by 2m... try it!

User Avatar

Wiki User

16y ago

Still curious? Ask our experts.

Chat with our AI personalities

LaoLao
The path is yours to walk; I am only here to hold up a mirror.
Chat with Lao
SteveSteve
Knowledge is a journey, you know? We'll get there.
Chat with Steve
TaigaTaiga
Every great hero faces trials, and you—yes, YOU—are no exception!
Chat with Taiga

Add your answer:

Earn +20 pts
Q: The sum of 25 consecutive integers is divisible by?
Write your answer...
Submit
Still have questions?
magnify glass
imp