philipssonicare wrote:
I tried to double-check when solving by using the manual way. 24 minus the factors
3, 5, 15, 25, 75
But this makes 19. I couldn't reconcile this for much too long.
Do not forget that 1 is a factor!
2^3* 3*5^2
factors of which can be written as 2^0 *(3*5^2) which is nothing but 1*(3*5^2)
2^1 *( 3*5^2)
2^2 *(3*5^2)
2^3*(3*5^2)
and for each of the above 3*5^2 will have (1+1)*(2+1) factors, so total number of factors is 6....
So for every (2^0, 2^1, 2^2 and 2^3) there are 6 combinations...
So for even factors at least one 2 should be present which gives 6*3 =18 factors
Now while calculating 6 factors for each 2 series we have already counted when each of them becomes 0
i.e Factors of 3*5^2 can be thought of (3^0*5^0=1, 3^0*5=5, 3^0*5^2=25, 3*5^0=3, 3*5^1=15, 3*5^2=75), so 6 factors are (1,5,25,3,15,75)
so 1 gets accounted in 6 factors of 3*5^2