Determine the set of natural numbers n such that every sum of n consecutive natural numbers is divisible by n?

Question

how do you find this out? =S

Answer 1

Nice question!!!!!

The answer is -- it's the ODD ones!

Here's why. The sum of any n consecutive natural numbers will be equal to the sum of the FIRST n natural numbers plus a multiple of n. (Convince yourself of that before going any further.)

And we know that the sum of the first n consecutive natural numbers = n(n+1)/2.

n divides that if and only if (n+1)/2 is an integer!

I.e., if and only if n is odd.

Q.E.D.

Answer 2

A sum of n consecutive natural numbers is an arithmetic progression with n terms, first term say m and last term m+n-1

The sum then is n/2 * ( m + (m+n-1) )

Simplify this and determine which n make this divisible by n.

Answer 3

it rather is quite ordinary to coach. If it have been real, than (n^5-n)/5 must be an integer huge style. additionally, the subsequent bigger n has to stick to this rule, for that reason ((n+a million)^5-(n+a million))/5 must be yet another integer huge style. for that reason, in turn, in case you subtract the decrease one (first one) from the better one, than you get yet another integer huge style. We only could coach that his is the case for all n. ordinary: ((n+a million)^5-(n+a million)-(n^5-n))/5 = ... (here we use the Pascal triangle for calculating the binomial coefficients) a million a million a million a million 2 a million a million 3 3 a million a million 4 6 4 a million a million 5 10 10 5 a million The n^5 term cancels, the a million term cancels too (a million-a million). for that reason we are left with 5*n^4+10*n^3+10*n^2+5n and dividing this with the aid of 5 leaves n^4+2*n^3+2*n^2+n and for that reason, for all n it rather is an integer.

Answer 4

Recall formula for sum of integers (due to Gauss)
Sum from k=1 to m of [k] =1+2+...m = m(m+1)/2
Then the sum of integers starting with m1+1 and ending with m2 is
S = 1+2+...m2 - (1+2+...m1)
S = m2(m2+1)/2-(m1+1)m1/2
S = [(m2)^2 - (m1)^2 + m2 -m1]/2
The number of integers is
n = m2 - m1
We want S/n to be an integer j, i.e.,
S/n = [(m2)^2 - (m1)^2 + m2 -m1]/2[m2 - m1] = j
which can be written as
(m2-m1)(m2+m1+1-2j)=0
Since m2 cannot equal m1, the only solution is
j = [m2 + (m1+1)]/2
Thus for j to be an integer the starting number (m1+1) and the ending number m2 must both be odd or both be even. This can be verified by examples:
(3+4+5)/3 = 4 integer - odd,odd
(3+4+5+6)/4 =9/2 non-integer - odd,even
(4+5+6)/3 = 5 integer - even,even