Problem 222
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See also the topics:
Don't post any spoilers
Comments, questions and clarifications about PE problems.
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ozgur
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Problem 222
My shortest pipe results in two intersecting spheres. One sphere (sn) in the middle intersects with the sphere two levels below it (sn-2). To account for that, I move sn along the pipe a little so that it does not intersect with sn-2. Am I on the right track, or should the correct solution result in no such intersections?
- ed_r
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axelbrz
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Re: Problem 222
Hi, can anyone confirm me if the length in micrometers of the shortest pipe, of internal radius 50mm, that can fully contain 3 balls of radii 48mm, 49mm and 50mm, is 293949?
Thank you!
Thank you!
"think(O(n))+O(n) sometimes is better than think(O(1))+O(1)"


- stijn263
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Re: Problem 222
Just putting the 3 balls on top of eachother you arrive at 294,000 micrometers. So 293,949 seems correct, but you should be able to verify this for yourself (by checking all 6 configurations)
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axelbrz
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Re: Problem 222
Yes, I did it! :) But I just wanted to know if I got the concept because my answer is wrong.
I'll check it.
Thanks!
I'll check it.
Thanks!
"think(O(n))+O(n) sometimes is better than think(O(1))+O(1)"


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axelbrz
- Posts: 51
- Joined: Mon Sep 08, 2008 5:34 am
Re: Problem 222
Mmm.. I'm also getting a wrong answer...
Can anyone confirm me if the shortest length of the pipe in micrometers for balls with radii 28, 31, 34, 35, 36, 37, 40, 45, 49 and 50 mm is 648649?
Thanks!
Can anyone confirm me if the shortest length of the pipe in micrometers for balls with radii 28, 31, 34, 35, 36, 37, 40, 45, 49 and 50 mm is 648649?
Thanks!
"think(O(n))+O(n) sometimes is better than think(O(1))+O(1)"


- daniel.is.fischer
- Posts: 2400
- Joined: Sun Sep 02, 2007 11:15 pm
- Location: Bremen, Germany
Re: Problem 222
No, it can't be that short.
Il faut respecter la montagne -- c'est pourquoi les gypaètes sont là.
- daniel.is.fischer
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Re: Problem 222
No, it's 293933.axelbrz wrote:Hi, can anyone confirm me if the length in micrometers of the shortest pipe, of internal radius 50mm, that can fully contain 3 balls of radii 48mm, 49mm and 50mm, is 293949?
Thank you!
Il faut respecter la montagne -- c'est pourquoi les gypaètes sont là.
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x10
- Posts: 8
- Joined: Sun Jul 19, 2009 8:20 pm
Re: Problem 222
I'm sorry to be reiterating the topic, but could anyone please post their answer for balls of sizes 40, 41... 50?
I get 983801.638816
I get 983801.638816

- daniel.is.fischer
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- Location: Bremen, Germany
Re: Problem 222
Too close to the problem's parameters to post the correct value, but yours is too high, the correct value is 981xxx.yyyyy
Il faut respecter la montagne -- c'est pourquoi les gypaètes sont là.
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mirzauzairbaig
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Re: Problem 222
Can anyone please list any particular basic topics that I need to go through to work on this problem ?
Last edited by mirzauzairbaig on Sat Mar 22, 2014 7:12 am, edited 1 time in total.
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TripleM
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Re: Problem 222
The best part about Project Euler is that you need to come up with the answers all by yourself, you won't be finding hints online 
Let's just say that a brute force approach doesn't need to check all N!/2 possibilities.
Let's just say that a brute force approach doesn't need to check all N!/2 possibilities.
- daniel.is.fischer
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- Joined: Sun Sep 02, 2007 11:15 pm
- Location: Bremen, Germany
Re: Problem 222
Let's say a brute force approach is supposed to not be a good idea. Look at the function that gives you the length or saved length locally. It has a property which lets you find the optimal arrangement quite easily (assuming what you're studying is mathematics or has a large mathematical component, like physics).
Il faut respecter la montagne -- c'est pourquoi les gypaètes sont là.
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TripleM
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jui-feng
- Posts: 11
- Joined: Thu Sep 02, 2010 5:20 pm
Re: Problem 222
It's kind of funny I was getting exactly the same (invalid) response for 40-50. It seems like we both made the same mistake that resulted in the same result, I only had some more decimal places (double precision).x10 wrote:I'm sorry to be reiterating the topic, but could anyone please post their answer for balls of sizes 40, 41... 50?
I get 983801.638816
Anyway, I solved it now.

- yourmaths
- Posts: 47
- Joined: Mon Aug 25, 2014 11:00 am
Re: Problem 222
One small quibble over the rounding of the final answer here. Let's say the answer to the problem is 123.4 micrometres. Rounding this (down since 0.4 < 0.5) we get 123 but this is no longer the shortest pipe since the sphere on top is now 0.4 micrometres above the top of the pipe. So my interpretation would be that the answer should be rounded up in all cases, giving 124, which was not accepted, but 123 was.
level = lambda number_solved: number_solved // 25


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DJohn
- Posts: 90
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Re: Problem 222
But that is not what the problem is asking for. What you're doing is finding the shortest integer length that satisfies the conditions. The problem is asking you to find the shortest length (which is not necessarily an integer) and then to round it to the nearest integer. That's a different thing. There's nothing in the problem statement that suggests that the rounded number has to meet any conditions other than being the closest integer to the shortest length.yourmaths wrote: Fri May 08, 2026 9:53 pm One small quibble over the rounding of the final answer here. Let's say the answer to the problem is 123.4 micrometres. Rounding this (down since 0.4 < 0.5) we get 123 but this is no longer the shortest pipe since the sphere on top is now 0.4 micrometres above the top of the pipe. So my interpretation would be that the answer should be rounded up in all cases, giving 124, which was not accepted, but 123 was.