Find-the-number-of-all-rational-numbers-m-n-such-that-i-0-lt-m-n-lt-1-ii-m-and-n-are-relatively-prime-and-iii-mn-25-

Question Number 13388 by Tinkutara last updated on 19/May/17

Find the number of all rational numbers

\frac{m}{n} such that (i) 0 < \frac{m}{n} < 1, (ii) m and

n are relatively prime and (iii) m n = 25! .

Commented by RasheedSindhi last updated on 19/May/17

There is at least one such number :

\frac{1}{25!}

Commented by prakash jain last updated on 19/May/17

25! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot \dots 25

= 2^{a} \cdot 3^{b} \cdot 5^{c} \cdot 7^{d} \cdot 11^{e} \cdot 13^{f} \cdot 17^{g} \cdot 19^{h} \cdot 23^{i}

Since m a n d n a r e c o p r i m e

o n l y w a y t o s e l e c t m i s c h o s i n g

a s u b s e t o f

{2, 3, 5, 7, 11, 13, 17, 19, 23}

= 2^{9} = 512

t o t a l s o l u t i o n s = \frac{512}{2} = 256

d i v i d e b y t w o s i n c e h a l f o f

c a s e s w i l l h a v e m a n d n s w a p p e d .

a n d o n l y o n e o f t h e m w i l l h a v e

m < n .

Commented by mrW1 last updated on 20/May/17

h o w i s t o u n d e r s t a n d t h e c o n d i c t i o n

t h a t m a n d n a r e r e l a t i v e l y p r i m e ?

Commented by prakash jain last updated on 20/May/17

Since m and n are relatively prime

and distinct prime factor of 25! are

{2, 3, 5, 7, 11, 13, 17, 19, 23}

we chose of subset of above set

say {2, 3, 5}

m = 2^{a} 3^{b} 5^{c}

n = 7^{d} 11^{e} 13^{f} 17^{g} 19^{h} 23^{i}

so m and n are relative prime .

but when we select

m = 7^{d} 11^{e} 13^{f} 17^{g} 19^{h} 23^{i}

n = 2^{a} 3^{b} 5^{c}

we discard one of the choice so divide by 2 .

Commented by mrW1 last updated on 20/May/17

T h a n k s!

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