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In how many ways 1500 be written as a product of three integers? Considering +ve and -ve integers.
It will be 124 ....
180 is the solution for non negative cases ...
however that is ordered ....
moreover there are solutions like (1*1*1500) or (2*2*375) which are having 2 integers common ... such arrangements are 4 in total ...
These account for (3!/2!) solutions each .. so 12 ..
now 168 remaining ordered non negative case has to be unordered ... by dividing by 3! cause each solution has occured for 3! times ...
so 168/6 = 28 ...
Now plug in the case for negative integers too ( say abc are the three integers ) then ( ab negative and C pos ) ( ac neg b pos) ( bc neg a pos) so 3 negative involvements
therefore total 4 such types ..
So 28*4+12 = 124
Hi Sahil,
Let the 3 numbers be x,y,z.
xyz = 1500 = 2^2*3*5^3
Let x = 2^a3^b5^c
y = 2^d 3^e 5^f
z = 2^g 3^h 5^i
Accordingly a+d+g = 2
b+e+h = 1
c+f+i = 3
Non negative integral solutions for these equations are 4C2*3C2*5C2 = 6*3*10 = 180
