In factored form, [imath]120=2^3\cdot 3\cdot 5[/imath] so there are [imath](3+1)(1+1)(1+1)=16[/imath] divisors.If "a" is a factor of 120, then find the no of positive integral solutions of X1.X2.X3=a
No of factors of 120 =16. (1,2,3,4,5,6,8,10...120)
Now how to proceed?
Therefore, a*b=120, call this equation "C", where both a and b are positive integers.If "a" is a factor of 120
Can you combine the equation in this quote with equation "C" to give something in a form that you already know how to solve?find the no of positive integral solutions of X1.X2.X3=a
The wording is odd. I suppose this means that x1 x2 x3 can be any divisor of 120. If so, you might just think of it as asking for the number of positive integer solutions of x1 x2 x3 x4 = 120. [Maybe this is the same idea Cubist has.]If "a" is a factor of 120, then find the no of positive integral solutions of X1.X2.X3=a
No of factors of 120 =16. (1,2,3,4,5,6,8,10...120)
Now how to proceed?
Nope, my idea was (a)*b=120 => (x1 x2 x3) b = 120just think of it as asking for the number of positive integer solutions of x1 x2 x3 x4 = 120. [Maybe this is the same idea Cubist has.]
120 divisors is irrelevant .Nope, my idea was (a)*b=120 => (x1 x2 x3) b = 120
Don't worry, I know this is actually the same as your suggestion. I'm impressed that you were able to see this directly using logical thinking!
So what have you tried at that point? This is your exercise to do, not ours.120 divisors is irrelevant .
a*b =120
Where I got stuck.
Otherwise the question is easy
Absolutely.Do you understand it? It's a simple extension of a problem we discussed before.
So you've solved it? Please show your work.Absolutely.
The same as previous one