All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
No way to run a railroad. (Posted on 2010-10-01) Difficulty: 3 of 5

BYE-LAWS of the H.PIPER STEAM-ENGINE RAILROAD (H.P.S-E.R.R.)

1. Every train on the H.P.S-E.R.R. must comprise only consecutively serial-numbered wagons.
2. Once chosen, such wagons must be assembled in ascending order of the quantity, size and (if necessary) power of the distinct prime factors of the wagons’ serial-numbers; e.g car 41 before car 1681(41,41) before car 39(3,13 )before car 30(2,3,5).
3. The lowest serial-numbered wagon must be in caboose* position.

If the serial-number of the first wagon in today's train is a perfect cube, what is the maximum length of the train?

(Small hint: wagon No. 194481 is the most recent addition to the rolling stock of the H.P.S-E.R.R.)
There is also a bonus, for the ingenious reader.

* a 'caboose' is the last wagon at the back end of the train.

No Solution Yet Submitted by broll    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
extra credit | Comment 3 of 5 |
(In reply to extra credit by Daniel)

Daniel,

Brilliant (and I was particularly gratified by the 'extra credit' answer)! As a pencil and paper person, I'm always profoundly impressed by the lightning speed with which which these responses are produced.

I have no comment to make on your solution, but have one point of clarification which I make by reference to the ordering of the last few numbers, i.e. those with 4 prime factors.You have:

4110 2*3*5*137
4170 2*3*5*139
4242 2*3*7*101
4134 2*3*13*53
4182 2*3*17*41
4218 2*3*19*37
4278 2*3*23*31
4230 2*3^2*5*47
4158 2*3^3*7*11
4130 2*5*7*79
4270 2*5*7*61
4186 2*7*13*23
4260 2^2*3*5*71
4140 2^2*3^2*5*23
4284 2^2*3^2*7*17
4180 2^2*5*11*19
4200 2^3*3*5^2*7
4095 3^2*5*7*13

whilst the actual order should be:

4200 2^3*3*5^2*7
4140 2^2*3^2*5*23
4230 2*3^2*5*47
4260 2^2*3*5*71
4110 2*3*5*137
4170 2*3*5*139
4158 2*3^3*7*11
4284 2^2*3^2*7*17
4242 2*3*7*101
4134 2*3*13*53
4182 2*3*17*41
4218 2*3*19*37
4278 2*3*23*31
4270 2*5*7*61
4130 2*5*7*79
4180 2^2*5*11*19
4186 2*7*13*23
4095 3^2*5*7*13

This is because the power of the prime factors only becomes relevant 'if necessary' i.e. as a 'tie-breaker' if all the prime factors would otherwise be identical. In the case of 4140 and 4230, for example, there is no need to consider the powers of 2 and 3, since the final prime factors, 23 and 47, can be used to determine the order of those two wagons.


Of course, this does not affect the order of the 'caboose' wagon, so does not invalidate the answer you have given

 

Edited on October 2, 2010, 3:44 pm
  Posted by broll on 2010-10-02 15:32:48

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information