Solution:

2+3+4=(1+2+3+4)-(1)

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

=(1+2+3+4+5+6+7+8+9)-(1+2+3+4+5+6+7+8)

This shows that the no of solutions to this problem is the no of tuples (n,m) such that

given number K can be expressed as K=(1+2+..........+n)-(1+2+3+...........+m)

and n > m

i.e K=n*(n+1)/2 -m*(m+1)/2

=(n-m)(n+m+1)/2

2*k=(n-m)*(n+m+1)

If we put the factor n-m=p and (n+m+1)=q,we get S=2*k=p*q

As p+q=2*n+1 is odd and S is even either n-m is odd otherwise n+m+1 is odd.

This leaves out with finding out the no of odd divisors of S,which can be easily done using prime factorization.

if S=(2^p)*(3^q)*(5^r)*(7^s)......

then the solution is (q+1)*(r+1)*(s+1)*.. i.e the no of odd divisors of S.

__click here for more questions__
## No comments:

## Post a Comment