Determine the total number of rational numbers of the form m/n, where m and n are positive integers such that:

(A) m/n lies in the interval (0, 1); and

(B) m and n are relatively prime; and

(C) mn = 25!

NOTE: "!" denotes the factorial symbol, where n! = 1*2*3*......*(n-1)*n

(In reply to Got it by Jer)

As there are 9 distinct prime factors of 25! -- 2, 3, 5, 7, 11, 13, 17, 19, and 23 -- and m and n are relatively prime, summing each number of combinations that form a fraction, m/n, to include the fractions with coprime value 1 as a numerator and denominator, will result in 512 rational numbers.  Exactly half of these will have a smaller numerator than the denominator, thus dividing 512 by 2 to give the result of 256 rational numbers where m/n lies in the interval (0,1). 

The formula used for each combination is P!/[r! * (P - r)!], where P is 9 and r represents one of the combinations to be summed. For example, where two prime factors are the factors of the numerator (the remaining seven prime factors are factors of the denominator) the number of combinations of prime factors that make up the numerator are 9!/[2!*(9-2)!].  Adding up each combination and dividing by 2 provides the result of 256.

*(This post was changed due to the error I made as pointed out by Charlie in the "following" post).

Edited on October 4, 2006, 2:50 pm

Posted by Dej Mar on 2006-10-03 15:01:39