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

Find the number of all rational numbers (m/n) such that (i) 0 < (m/n) < 1, (ii) m and n are relatively prime and (iii) mn = 25!.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{all}\:\mathrm{rational}\:\mathrm{numbers} \\ $$ $$\frac{{m}}{{n}}\:\mathrm{such}\:\mathrm{that}\:\left(\mathrm{i}\right)\:\mathrm{0}\:<\:\frac{{m}}{{n}}\:<\:\mathrm{1},\:\left(\mathrm{ii}\right)\:{m}\:\mathrm{and} \\ $$ $${n}\:\mathrm{are}\:\mathrm{relatively}\:\mathrm{prime}\:\mathrm{and}\:\left(\mathrm{iii}\right)\:{mn}\:=\:\mathrm{25}!. \\ $$

Commented byRasheedSindhi last updated on 19/May/17

There is at least one such number: (1/(25!))

$$\mathrm{There}\:\mathrm{is}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{such}\:\mathrm{number}: \\ $$ $$\frac{\mathrm{1}}{\mathrm{25}!} \\ $$

Commented byprakash 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.

$$\mathrm{25}!=\mathrm{1}\centerdot\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}\centerdot...\mathrm{25} \\ $$ $$=\mathrm{2}^{{a}} \centerdot\mathrm{3}^{{b}} \centerdot\mathrm{5}^{{c}} \centerdot\mathrm{7}^{{d}} \centerdot\mathrm{11}^{{e}} \centerdot\mathrm{13}^{{f}} \centerdot\mathrm{17}^{{g}} \centerdot\mathrm{19}^{{h}} \centerdot\mathrm{23}^{{i}} \\ $$ $$\mathrm{Since}\:{m}\:{and}\:{n}\:{are}\:{coprime} \\ $$ $${only}\:{way}\:{to}\:{select}\:{m}\:{is}\:{chosing} \\ $$ $${a}\:{subset}\:{of}\: \\ $$ $$\left\{\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{7},\mathrm{11},\mathrm{13},\mathrm{17},\mathrm{19},\mathrm{23}\right\} \\ $$ $$=\mathrm{2}^{\mathrm{9}} =\mathrm{512} \\ $$ $${total}\:{solutions}=\frac{\mathrm{512}}{\mathrm{2}}=\mathrm{256} \\ $$ $${divide}\:{by}\:{two}\:{since}\:{half}\:{of} \\ $$ $${cases}\:{will}\:{have}\:{m}\:{and}\:{n}\:{swapped}. \\ $$ $${and}\:{only}\:{one}\:{of}\:{them}\:{will}\:{have} \\ $$ $${m}<{n}. \\ $$

Commented bymrW1 last updated on 20/May/17

how is to understand the condiction that m and n are relatively prime?

$${how}\:{is}\:{to}\:{understand}\:{the}\:{condiction} \\ $$ $${that}\:{m}\:{and}\:{n}\:{are}\:{relatively}\:{prime}? \\ $$

Commented byprakash 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.

$$\mathrm{Since}\:{m}\:\mathrm{and}\:{n}\:\mathrm{are}\:\mathrm{relatively}\:\mathrm{prime} \\ $$ $$\mathrm{and}\:\mathrm{distinct}\:\mathrm{prime}\:\mathrm{factor}\:\mathrm{of}\:\mathrm{25}!\:\mathrm{are} \\ $$ $$\left\{\mathrm{2},\:\mathrm{3},\:\mathrm{5},\:\mathrm{7},\:\mathrm{11},\:\mathrm{13},\:\mathrm{17},\:\mathrm{19},\:\mathrm{23}\right\} \\ $$ $$\mathrm{we}\:\mathrm{chose}\:\mathrm{of}\:\mathrm{subset}\:\mathrm{of}\:\mathrm{above}\:\mathrm{set} \\ $$ $$\mathrm{say}\:\left\{\mathrm{2},\mathrm{3},\mathrm{5}\right\} \\ $$ $${m}=\mathrm{2}^{{a}} \mathrm{3}^{{b}} \mathrm{5}^{{c}} \\ $$ $${n}=\mathrm{7}^{{d}} \mathrm{11}^{{e}} \mathrm{13}^{{f}} \mathrm{17}^{{g}} \mathrm{19}^{{h}} \mathrm{23}^{{i}} \\ $$ $$\mathrm{so}\:{m}\:\mathrm{and}\:{n}\:\mathrm{are}\:\mathrm{relative}\:\mathrm{prime}. \\ $$ $$\mathrm{but}\:\mathrm{when}\:\mathrm{we}\:\mathrm{select} \\ $$ $${m}=\mathrm{7}^{{d}} \mathrm{11}^{{e}} \mathrm{13}^{{f}} \mathrm{17}^{{g}} \mathrm{19}^{{h}} \mathrm{23}^{{i}} \\ $$ $${n}=\mathrm{2}^{{a}} \mathrm{3}^{{b}} \mathrm{5}^{{c}} \\ $$ $$\mathrm{we}\:\mathrm{discard}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{choice}\:\mathrm{so}\:\mathrm{divide}\:\mathrm{by}\:\mathrm{2}. \\ $$

Commented bymrW1 last updated on 20/May/17

Thanks!

$${Thanks}! \\ $$

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