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!
12 and 13 :)
The integers are 24, 25 and 26.
25 and 26
42 + 52 = 16 + 25 = 41
That isn't possible. The three consecutive number are assumed to be integers; the sum of three consecutive integers is always a multiple of 3 (try it out).
They are odd consecutive integers: 21, 23, 25 and 27.
They are 24, 25 and 26, as well as the consecutive odd integers 23, 25 and 27.
12 and 13 :)
The integers are 24, 25 and 26.
24 + 25 + 26 = 75
-13, -12
25 and 26
25, 26 and 27
The three integers are -27, -26 and -25.
25, 26 and 27.
Two consecutive even integers will be one more and one less than the quotient of their sum divided by 2. The given sum of the two consecutive integers divided by 2 is 25, so the two consecutive integers are 24 and 26.
The numbers are 23, 24 and 25.