Number System : 112

4.In how many ways can 2004 be written as a sum of two or more consecutive positive integers? Ans 3


SOLUTION:  2004 = 4*3*167

The numbers summing up to 2004, can be an A.P with even number of terms or odd number of terms.

(i)A.P with ODD number of terms:

If an A.P has "odd number of terms", then the "middle term" is the arithmetic mean and the sum of all the terms = number of terms * middle term. [ A.P = ….,a-2d, a-d,a,a+d,a+2d,….]
So, to form an A.P of "consecutive" numbers summing up to 2004 [ A.P = ….,a-2, a-1,a,a+1,a+2,….] , we have to see the number of odd divisors of 2004 (since the number of terms has to be ODD).

The number different odd divisors of 2004 = 3,167,(3*167) = "3"
But, the series of consecutive terms of an A.P 
with n= 167 and middle term = 12 - will HAVE NEGATIVE NUMBERS,
with n= 501 and middle term = 4 - will HAVE NEGATIVE NUMBERS,


so, ONLY '1' POSSIBLE CASE, 


n= 3 and middle term = 668 - CAN ADD UP TO 2004.

(ii)A.P with EVEN number of terms:

If an A.P has "even number of terms", then the "average of the middle terms" is the arithmetic mean and 
the sum of all the terms = number of terms * average of the middle terms. [ A.P = ….,a-2d, a-d,'a,a+d',a+2d,….]
So, to form an A.P of "consecutive" numbers summing up to 2004 [ A.P = ….,a-2, a-1,a,a+1,a+2,….] , we have to see the number of odd divisors of 2004 (since the average of any 2 consecutive number is ALWAYS ODD).
The number different odd divisors of 2004 = 3,167,(3*167) = "3"
But, the series of consecutive terms of an A.P 
with n=4 and avg. of middle term = 501 - will HAVE NEGATIVE NUMBERS,


so, there are ONLY '2' POSSIBLE CASES, 

(a)n= 12 and avg. of middle term = 167 - CAN ADD UP TO 2004.

(b)n= 4 and avg. of middle term = 501 - CAN ADD UP TO 2004


HENCE, BY COMBINING THE RESULTS OF (i) and (ii),
2004 be written as a sum of two or more consecutive positive integers in '3' ways.