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| A rational number problem (Posted on 2006-10-02) |
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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
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Submitted by K Sengupta
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Rating: 5.0000 (1 votes)
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Solution:
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(Hide)
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We have:
25! = (2^m)*(3^n)*(5^k)*(7^r)*(11^t)*13*17*19*23...........(1)
Let S denote the set {2^m, 3^n, 5^k, ......., 23}consisting of the nine prime factors. this posssesses 2^9 subsets including the empty set. If A belongs to S and A is not empty, let the product of the members of A be a. If A is empty, then we choose a =1. The fundamental theorem of arithmetic shows that distinct subsets will give rise to distinct values of a. Thus we obtain 2^9 members of a.
If b = 25!/a; then ab = 25! and a and b are relatively prime. it is clear from (1) that 25! is not a perfect square; so that, a is not equal to b. If a is less than b, we choose m = a and n = b and, if a is greater than b we choose m =b and n=a. Every fraction m/n of the desired type is obtained like this. Moreover, each m/n is obtained exactly twice. Since, we have 2^9 pairs (a,b); the required number of fractions of the desired type is 1/2*(2^9)= 256.
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