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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
$25!=1∙2∙3∙4∙...25 =2^a ∙3^b ∙5^c ∙7^d ∙11^e ∙13^f ∙17^g ∙19^h ∙23^i Since m and n are coprime only way to select m is chosing a subset of {2,3,5,7,11,13,17,19,23} =2^9 =512 total solutions=((512)/2)=256 divide by two since half of cases will have m and n swapped. and only one of them will have m<n.$
$25! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot . . . 25$ $= 2^{a} \cdot 3^{b} \cdot 5^{c} \cdot 7^{d} \cdot 11^{e} \cdot 13^{f} \cdot 17^{g} \cdot 19^{h} \cdot 23^{i}$ $Since m a n d n a r e c o p r i m e$ $o n l y w a y t o s e l e c t m i s c h o s i n g$ $a s u b s e t o f$ ${2, 3, 5, 7, 11, 13, 17, 19, 23}$ $= 2^{9} = 512$ $t o t a l s o l u t i o n s = \frac{512}{2} = 256$ $d i v i d e b y t w o s i n c e h a l f o f$ $c a s e s w i l l h a v e m a n d n s w a p p e d .$ $a n d o n l y o n e o f t h e m w i l l h a v e$ $m < n .$
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
$Since m and n are relatively prime and distinct prime factor of 25! are {2, 3, 5, 7, 11, 13, 17, 19, 23} we chose of subset of above set say {2,3,5} m=2^a 3^b 5^c n=7^d 11^e 13^f 17^g 19^h 23^i so m and n are relative prime. but when we select m=7^d 11^e 13^f 17^g 19^h 23^i n=2^a 3^b 5^c we discard one of the choice so divide by 2.$
$Since m and n are relatively prime$ $and distinct prime factor of 25! are$ ${2, 3, 5, 7, 11, 13, 17, 19, 23}$ $we chose of subset of above set$ $say {2, 3, 5}$ $m = 2^{a} 3^{b} 5^{c}$ $n = 7^{d} 11^{e} 13^{f} 17^{g} 19^{h} 23^{i}$ $so m and n are relative prime .$ $but when we select$ $m = 7^{d} 11^{e} 13^{f} 17^{g} 19^{h} 23^{i}$ $n = 2^{a} 3^{b} 5^{c}$ $we discard one of the choice so divide by 2 .$
Commented by mrW1 last updated on 20/May/17

$T h a n k s!$
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