Problem 135

A place to air possible concerns or difficulties in understanding ProjectEuler problems. This forum is not meant to publish solutions. This forum is NOT meant to discuss solution methods or giving hints how a problem can be solved.
Forum rules
As your posts will be visible to the general public you
are requested to be thoughtful in not posting anything
that might explicitly give away how to solve a particular problem.

This forum is NOT meant to discuss solution methods for a problem.

In particular don't post any code fragments or results.

Don't start begging others to give partial answers to problems

Don't ask for hints how to solve a problem

Don't start a new topic for a problem if there already exists one


See also the topics:
Don't post any spoilers
Comments, questions and clarifications about PE problems.
TripleM
Posts: 382
Joined: Fri Sep 12, 2008 3:31 am

Re: Problem 135

Post by TripleM »

No, there isn't.

JMW1994
Posts: 43
Joined: Sat Apr 09, 2011 11:35 pm

Re: Problem 135

Post by JMW1994 »

Thanks for the clarification.
Image

Kevin_L
Posts: 5
Joined: Sun Apr 01, 2012 1:18 pm

Re: Problem 135

Post by Kevin_L »

A friend of mine spoiled the answer to this one for me, so I know my current wrong answer is close (my answer is 12 too small), but I refuse to type in the correct answer until I get it myself.

Can anyone confirm or deny the following partial results:

Let d be the common difference of the arithmetic progression in question. If you restrict yourself to arithmetic progressions that have d<=400, 550, 10^3, or 10^4, then the last three digits of the answers would be
Expand
419, 770, 462, 970
respectively.

It's killing me trying to find out where the missing 12 solutions are.

alice0meta
Posts: 1
Joined: Mon Aug 05, 2013 8:31 pm

Re: Problem 135

Post by alice0meta »

Kevin_L wrote:A friend of mine spoiled the answer to this one for me, so I know my current wrong answer is close (my answer is 12 too small), but I refuse to type in the correct answer until I get it myself.

Can anyone confirm or deny the following partial results:

Let d be the common difference of the arithmetic progression in question. If you restrict yourself to arithmetic progressions that have d<=400, 550, 10^3, or 10^4, then the last three digits of the answers would be
Expand
419, 770, 462, 970
respectively.

It's killing me trying to find out where the missing 12 solutions are.
you're letting z equal 0

z != 0

Neilius
Posts: 8
Joined: Mon Aug 24, 2015 5:27 am
Contact:

Re: Problem 135

Post by Neilius »

I am enjoying this problem very much.
I would like to verify my algorithm for a values of n other than 1155.
The next higher value of n I get with ten solutions is n=1755.
I suspect this is incorrect.
Am I incorrect in my calculation that n=1755 has ten solutions?
Does it really have more?
Thanks in advance.
Image

User avatar
dawghaus4
Posts: 54
Joined: Fri Nov 29, 2013 2:22 am

Re: Problem 135

Post by dawghaus4 »

Neilius wrote:..
The next higher value of n I get with ten solutions is n=1755.
I suspect this is incorrect.
...
It is you suspicion that is incorrect.

Tom

Neilius
Posts: 8
Joined: Mon Aug 24, 2015 5:27 am
Contact:

Re: Problem 135

Post by Neilius »

dawghaus4 wrote:
It is you suspicion that is incorrect.

Tom
Many thanks.
I finally worked it out.
Program runs in about 6 seconds.
A very enjoyable problem :)
Image

DanielJackson
Posts: 2
Joined: Wed Feb 19, 2020 11:03 pm

Re: Problem 135

Post by DanielJackson »

n = a*b, a > b
Let n = 15 = 15*1 = 5*3
1. a = 15, b = 1 => d = 4, 19^2 - 15^2 - 11^2 = 15
2. a = 5, b = 3 => d = 2, 7^2 - 5^2 - 3^2 = 15
n = 15 has exactly 2 solutions and 15 < 27 (by condition). Where am I wrong?

mdean
Posts: 156
Joined: Tue Aug 02, 2011 2:05 am

Re: Problem 135

Post by mdean »

DanielJackson wrote:
Wed Feb 19, 2020 11:28 pm
n = a*b, a > b
Let n = 15 = 15*1 = 5*3
1. a = 15, b = 1 => d = 4, 19^2 - 15^2 - 11^2 = 15
2. a = 5, b = 3 => d = 2, 7^2 - 5^2 - 3^2 = 15
n = 15 has exactly 2 solutions and 15 < 27 (by condition). Where am I wrong?
$5^2-3^2-1^2=15$. 15 has at least 3 solutions.
Image

DanielJackson
Posts: 2
Joined: Wed Feb 19, 2020 11:03 pm

Re: Problem 135

Post by DanielJackson »

mdean wrote:
Thu Feb 20, 2020 12:40 am
$5^2-3^2-1^2=15$. 15 has at least 3 solutions.
I realized I hadn't considered it. Thanks

Post Reply