How many factors are there for X = 2^7 * 5^4 * 7^8 which are neither the factors of Y = 2^3 * 5^2 * 7^5 nor of Z = 2^5 * 5^3 * 7^3

Hi
78 = 2*3*13
So x = 2^8*3*5^4*13
If the powers of 2 are 6 or 7 or 8, then the factors won't be common between the first number and the other two numbers. Such cases would be 3*2*5*2 = 60
Similarly if the power of 5 is 4, there will not be common factors. However we need to keep in mind that we have already counted the cases when 2 is 6,7,8 and 5 is 4. So we should not count these cases.
Hence required factors would be 5*2*1*2 = 20
Similarly, for 3, we will have 5*4*2 = 40
The cases where 13 is present have already been counted so we should not count them again. Thus, the required number of cases would be 60 + 20 + 40 = 120
Hope that helps.