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
The rest by Larry Settle)
These are your proposed 16 fractions:
rational factored
1 / 15511210043330985984000000; 1 / 23 19 17 13 R
23 / 674400436666564608000000; 23 / 19 17 13 R
19 / 816379475964788736000000; 19 / 23 17 13 R
437 / 35494759824556032000000; 23 19 / 17 13 R
17 / 912424120195940352000000; 17 / 23 19 13 R
391 / 39670613921562624000000; 23 17 / 19 13 R
323 / 48022322115575808000000; 19 17 / 23 13 R
7429 / 2087927048503296000000; 23 19 17 / 13 R
13 / 1193170003333152768000000; 13 / 23 19 17 R
299 / 51876956666658816000000; 23 13 / 19 17 R
247 / 62798421228060672000000; 19 13 / 23 17 R
5681 / 2730366140350464000000; 23 19 13 / 17 R
221 / 70186470784303104000000; 17 13 / 23 19 R
5083 / 3051585686274048000000; 23 17 13 / 19 R
4199 / 3694024778121216000000; 19 17 13 / 23 R
96577 / 160609772961792000000; 23 19 17 13 /R
You ask to "try to move any factor in R from n to m." Note you rightly did not say prime factor, so multiple prime factors can be moved. Let's move 2^22 from n to m. The last fraction above becomes 405073297408 / 38292353859375; and m and n are still relatively prime. The same could have been done with 11^2 instead, or with both 2^22 and 11^2, etc., just so long as you move all the 2's together, all the 11's together, all the 3's together. And if you move so many that m becomes bigger than n, then just reverse them; that's the reason for the original division by 2. The 2^22 acts as a unit, as does the 3^10 and the 5^6, etc. There's no reason for the 2's to be on the same side as the 5's, or the 3's on the same side as the 11's, etc.
Edited on October 7, 2006, 12:43 am
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Posted by Charlie
on 2006-10-07 00:27:02 |
re: The rest
