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Problem 135 is providing wrong informatio
Posted: Mon Dec 24, 2007 11:42 am
by Marcscol
Hi Euler,
I was trying to solve problem 135 at it says that the least value that an equation has exactly two solutions is 27:
34^2 - 27^2 - 20^2 = 12^2 - 9^2 - 6^2 = 27
Trying to solve it I got that 15 has two solutions also:
19^2 - 15^2 - 11^2 = 7^2 - 5^2 - 3^2 = 15
Maybe I'm wrong, maybe I'm doing something wrong, but this is the thing that I've noticed.
Thanks and Regards,
Marcus
Re: Problem 135 is providing wrong informatio
Posted: Mon Dec 24, 2007 1:22 pm
by xenon
How about 5^2 -3^2 -1^2?
Re: Problem 135 is providing wrong informatio
Posted: Mon Dec 24, 2007 5:20 pm
by Marcscol
Yeah.... After posting, I checked the algo and I saw that I was making a mistake.
anyway, thanks for the reply.
Marcus
Problem 135
Posted: Tue Nov 11, 2008 5:53 pm
by GoSlow2GoFast
I've been looking at problem 135, and seem to be missing something. It states that for n=27, there are only 2 solutions:
342 - 272 - 202 = 27
122 - 92 - 62 = 27
Why aren't others included, such as:
82 - 62 - 12 = 27
102 - 82 - 32 = 27
142 - 122 - 52 = 27
162 - 152 - 22 = 27
I must be missing something in the wording of this problem, but it's not jumping out at me.
~gs2gf
Re: Am I reading problem 135 wrong?
Posted: Tue Nov 11, 2008 6:07 pm
by quilan
The numbers have to be part of an arithmetic progression.
i.e. (x+2k)^2 - (x+k)^2 - x^2 = n
Re: Am I reading problem 135 wrong?
Posted: Tue Nov 11, 2008 6:12 pm
by GoSlow2GoFast
Duh, I was completely "reading over" that clause and it just wasn't being noticed. Thanks.
~gs2gf
Re: Problem 135
Posted: Sat Jul 10, 2010 12:11 am
by MaJJ
If I use quilan's words -
- then my loop stops when k = 578 and
Expand
605 values of n have 10 solutions
. Is the latter way off?
The
k gives sense:
(1+578+578)^2 - (1+578)^2 - 1^2 = 1338649 - 335241 - 1 = 1003407
(1+577+577)^2 - (1+577)^2 - 1^2 = 1334025 - 334084 - 1 = 999940
- or am I missing something?
Re: Problem 135
Posted: Sat Jul 10, 2010 12:57 am
by TripleM
I'm afraid your answer isn't close.
Re: Problem 135
Posted: Thu Jul 07, 2011 8:33 am
by kapila_619
I don't know whether this is a valid question
I think I'm getting my maths wrong somewhere
to find, values of z for which (z+2k)^2-(z+k)^2-z^2=n
f(z,k)=z^2-2zk-3k^2+n //n is const
I have found condition for z>0 to be : k belongs to -sqrt(n)/2 to sqrt(n)/2
but
n=27,x=7,y=20
and
n=27,x=3,y=6
these don't satisfy my obtained conditioned, while they satisfy f(z,k)=0
for n=27, k should belong to -2.598 to 2.598
but k=3,7 satisfy
I know we have to find f(z,k)=0, but the problem says x,y,z>0, so i just applied the condition discriminant<0 to get a range of values of 'k' for a particular 'n'
can u tell me where am I going wrong
thanks
Re: Problem 135
Posted: Thu Jul 07, 2011 9:03 am
by jaap
kapila_619 wrote:I have found condition for z>0 to be : k belongs to -sqrt(n)/2 to sqrt(n)/2
I think you have a sign error on the k^2 term of the discriminant. The one I get is always positive (for positive n) so of no use.
Edit: No, I got the sign wrong myself. So you get an expression in k and n for the discriminant. You need that discriminant to be positive for there to be any solutions for z at all. So you need k to lie outside the region you found rather than inside.
Re: Problem 135
Posted: Mon Oct 03, 2011 6:36 pm
by lirugeda
I found an algorithm to get the answer, but when I put my answer in, it's incorrect. So, I was wondering if someone could help me verify the first 10 values of n which have exactly 10 distinct solutions:
1155
1995
2415
3003
3135
3255
3315
3795
3927
4095
Re: Problem 135
Posted: Mon Oct 03, 2011 7:07 pm
by jaap
The correct ones are 1155, 3003, 3135 (which are first, 7th and 8th on my list).
Re: Problem 135
Posted: Tue Oct 04, 2011 4:48 pm
by xe3tec
are these the first 3?
[spoiler]1155
1184
1203[/spoiler]
Re: Problem 135
Posted: Tue Oct 04, 2011 5:16 pm
by jaap
No. The first one is correct, but the others don't have 10 solutions but only two.
Re: Problem 135
Posted: Tue Oct 04, 2011 6:45 pm
by xe3tec
edit: ok thx
I am stuck
Re: Problem 135
Posted: Wed Oct 05, 2011 10:51 am
by xe3tec
[spoiler]1155
[1155, 1575, 1755, 1920, 2079, 2304, 2475, 2688, 2835, 2880, 3003][/spoiler]
which one of these are wrong?
Re: Problem 135
Posted: Sun Jun 24, 2012 2:20 am
by Tharsis
jaap wrote:The correct ones are 1155, 3003, 3135 (which are first, 7th and 8th on my list).
Actually, 3315 is also correct.
xe3tec wrote:[spoiler]1155
[1155, 1575, 1755, 1920, 2079, 2304, 2475, 2688, 2835, 2880, 3003][/spoiler]
which one of these are wrong?
1575, 2475, 2835, and 2880 are not correct.
Re: Problem 135
Posted: Thu Dec 13, 2012 2:10 pm
by thkang
Given the positive integers, x, y, and z, are consecutive terms of an arithmetic progression, the least value of the positive integer, n, for which the equation, x^2 -y^2- z^2 = n, has exactly two solutions is n = 27:
34^2- 27^2- 20^2 = 12^2- 9^2-6^2 = 27
It turns out that n = 1155 is the least value which has exactly ten solutions.
How many values of n less than one million have exactly ten distinct solutions?
I'm having problem with counting N:
let's say we're thinking about arithmetic progression that progresses by 10. since 0<n<1,000,000, here are all the possible cases:
Code: Select all
P|||| X Y Z N
p==10 21 11 1 319
p==10 22 12 2 336
p==10 23 13 3 351
p==10 24 14 4 364
p==10 25 15 5 375
p==10 26 16 6 384
p==10 27 17 7 391
p==10 28 18 8 396
p==10 29 19 9 399
p==10 30 20 10 400
p==10 31 21 11 399
p==10 32 22 12 396
p==10 33 23 13 391
p==10 34 24 14 384
p==10 35 25 15 375
p==10 36 26 16 364
p==10 37 27 17 351
p==10 38 28 18 336
p==10 39 29 19 319
p==10 40 30 20 300
p==10 41 31 21 279
p==10 42 32 22 256
p==10 43 33 23 231
p==10 44 34 24 204
p==10 45 35 25 175
p==10 46 36 26 144
p==10 47 37 27 111
p==10 48 38 28 76
p==10 49 39 29 39
now the problem is I'm having hard time understanding what
n I should count.
the least value of the positive integer, n,
I can see that 49, 39, 29 yields least value of n, which is 39.
(and this is trivial without enumerating all x, y, z)
initially, given this observation, I thought what I had to do was:
for p==10, the least value of n is 39, count it once, move along to next p
however, by this method the given example of n=1155 having 10 solutions could not be found.
edit :
disregard everything I wrote.
however I have a question : how could n have a 'least' value? doesn't that term implies there is also a 'most' value and there are more than one possible 'n's for an arithmetic sequence?
Re: Problem 135
Posted: Thu Dec 13, 2012 6:15 pm
by TheEvil
thkang wrote:
however I have a question : how could n have a 'least' value? doesn't that term implies there is also a 'most' value and there are more than one possible 'n's for an arithmetic sequence?
If I remember well, in most of the cases least means the smallest number.
By this I mean: let us fix an arbitrary n which is less than 1155. The number of different triplets (x,y,z) for which x^2-y^2-z^2=n for this fixed n, is at most nine, or at least eleven. For n=1155, the number of triplets is exactly ten.
Re: Problem 135
Posted: Sun Feb 17, 2013 9:52 pm
by JMW1994
On this problem, is there a difference between "exactly ten solutions" and "exactly ten distinct solutions?" I found this very confusing since my code is able to get jaap's numbers in the correct order and yet my answer is wrong.