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Topic: a 6-digit number (Read 1069 times) |
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Benny
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a 6-digit number
« on: Jan 2nd, 2010, 10:14am » |
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Find, showing your method, a six-digit integer N with the following properties:
- N is a perfect square
- the number formed by the last three digits of N is exactly one greater than the number formed by the first three digits of N.
For example, the number N might look like 123124.
However, it is not a square.
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ThudnBlunder
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Re: a 6-digit number
« Reply #1 on: Jan 2nd, 2010, 11:09am » |
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Have you found such a number or is it just something you have always wondered about? 
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Benny
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Re: a 6-digit number
« Reply #2 on: Jan 2nd, 2010, 11:14am » |
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on Jan 2nd, 2010, 11:09am, ThudanBlunder wrote:| Have you found such a number or is it just something you have always wondered about? |
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No, I haven't found one, but I'm still looking ... I need help.
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Benny
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Re: a 6-digit number
« Reply #3 on: Jan 2nd, 2010, 12:33pm » |
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I found 4282 = 183184
I went to check the list of all 6-digit numbers that are squares
Here's a link: http://www.naturalnumbers.org/PS-1000.txt
Could anyone find an elegant method to find such a number (a mathematical one, not programming)?
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| « Last Edit: Jan 2nd, 2010, 5:44pm by Benny » |
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JohanC
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Re: a 6-digit number
« Reply #4 on: Jan 2nd, 2010, 1:31pm » |
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on Jan 2nd, 2010, 12:33pm, BenVitale wrote:| Could anyone find an elegant method to find such a number (a mathematical one, not programming)? |
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Do you have any indication to suspect that such a mathematical derivation should exist?
Checking them one by one might be simpler:
8281=912
183184=4282
328329=5732
528529=7272
715716=8462
60996100=78102
82428241=90792
98029801=99012
1322413225=363652
4049540496=636362
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Benny
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Re: a 6-digit number
« Reply #5 on: Jan 2nd, 2010, 2:02pm » |
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on Jan 2nd, 2010, 1:31pm, JohanC wrote:
Do you have any indication to suspect that such a mathematical derivation should exist? |
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I'm not sure if there's one.
some of the numbers you posted are not 6-digit numbers.
I know that the number N is
100489 < N < 998001
317^2 < N < 999^2
the answer being 428^2 = 183184
N = abcdef,
if we set A = abc and B = def
f - c = 1
A = c + 10b + 100a
B = f + 10e + 100d
a = d, b = e
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| « Last Edit: Jan 2nd, 2010, 5:45pm by Benny » |
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towr
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Re: a 6-digit number
« Reply #6 on: Jan 2nd, 2010, 3:18pm » |
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I seem to be missing 1 from the list Johan gave, and it's too late to consider what I messed up.
Here's my derivation for 3 such 6 digit numbers, anyway.
1001*x+1=n2 with 100 <= x <=998
1001*x = (n+1)(n-1)
7*11*13*x = (n+1)(n-1)
(n+1)=7*11*a & (n-1)=13*b
(n+1)=7*13*a & (n-1)=11*b
(n+1)=11*13*a & (n-1)=7*b
13b+2=77a -> 2 = 12a (mod 13) -> a=11 (mod 13)
11b+2=91a -> 2 = 3a (mod 11) -> a=8 (mod 11)
7b+2=143a -> 2 = 3a (mod 7) -> a=3 (mod 7)
13b +2 = 77(13k+11) -> b=77k+65
11b +2 = 91(11k+8) -> b=91k+66
7b +2 = 143(7k+3) -> b=143k+61
316 < n < 1000
316 <13b +1 < 1000 => 24 < b < 77 => (13*65+2)(13*65) = 715715
316 <11b +1 < 1000 => 18 < b < 91 => (11*66+2)(11*66) = 528528
316 <7b +1 < 1000 => 44 < b < 143 => (7*61+2)(7*61) = 183183
(And remember, these are n2-1)
[edit]Oh wait, I see what I forgot; I only considered cases where (n+1) contains 2 of the factors of 1001.[/edit]
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| « Last Edit: Jan 2nd, 2010, 3:20pm by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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ThudnBlunder
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Re: a 6-digit number
« Reply #7 on: Jan 2nd, 2010, 3:46pm » |
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We have 1001y = (x + 1)(x - 1) for some x,y
So y = (x + 1)(x - 1)/(7*11*13)
There are only 6 ways that the 3 factors 7,11,13 can divide into (x + 1)(x - 1), eg. 7 and 11 might divide into x + 1 and 13 into x - 1
This leads to the following solvable Diophantine equations:
7k = 143m  2
11k = 91m 2
13k = 77m 2
The first equation gives k = 61, m = 3 and k = 82, m = 4
leading to x = 428, y= 183 and x = 573, y = 328
The second equation gives k = 66, m= 8
leading to x = 727, y = 528
The third equation gives k = 65, m = 11
leading to x = 846, y = 715
Any other solutions are out of range.
Edit: Shucks, towr beat me to it, albeit incompletely.
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| « Last Edit: Jan 4th, 2010, 11:25am by ThudnBlunder » |
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Benny
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Re: a 6-digit number
« Reply #8 on: Jan 2nd, 2010, 6:48pm » |
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on Jan 2nd, 2010, 3:46pm, ThudanBlunder wrote:We have 1001y = (x + 1)(x - 1) for some x,y
So y = (x + 1)(x - 1)/(7*11*13)
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I think I got it, now.
1001y + 1 = x^2
Now, if I want to find all the 6-digit numbers such as the number formed by the last three digits of N is exactly a square number greater than the number formed by the first three digits of N.
e.g. 4, 9, ..., 64, 81, 100, ..., 256, 289, ..., 400
I would need to write:
1001y + 4 = x^2 or 1001y = x^2 - 2^2 = (x-2)(x+2)
In order to find the number 548^2 = 300304, the difference is 4 (2^2)
..............
1001y + 64 = x^2 or 1001y = x^2 - 8^2
in order to find the number 580^2 = 336400, the difference is 64 (8^2)
1001y + 100 = x^2 or 1001y = x^2 - 10^2
the numbers are 452^2 = 204304, 549^2 = 301401
1001y + 289 = x^2 or 1001y = x^2 - 17^2
the number is 633^2 = 400689
1001y + 400 = x^2 or 1001y = x^2 - 20^2
the numbers are 475^2 = 225625, 526^2 = 276676, 552^2 = 304704.
etc.
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| « Last Edit: Jan 2nd, 2010, 6:50pm by Benny » |
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ThudnBlunder
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Re: a 6-digit number
« Reply #9 on: Jan 3rd, 2010, 5:41am » |
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How are you finding your numbers?
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Benny
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Re: a 6-digit number
« Reply #10 on: Jan 3rd, 2010, 10:27am » |
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on Jan 3rd, 2010, 5:41am, ThudanBlunder wrote:| How are you finding your numbers? |
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I asked myself, what was your method to find the number if the difference is one. You wrote:
We have 1001y = (x + 1)(x - 1) for some x,y
or 1001y + 1 = x^2
For example,
I asked myself : can I find a number such as the difference is 4
If there's one then I could write:
1001y + 4 = x^2
............
etc.
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ThudnBlunder
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Re: a 6-digit number
« Reply #11 on: Jan 3rd, 2010, 10:36am » |
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on Jan 3rd, 2010, 10:27am, BenVitale wrote:
If there's one then I could write:
1001y + 4 = x^2 |
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I mean, from 1001y = (x + 2)(x - 2) are you using trial and error?
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Benny
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Re: a 6-digit number
« Reply #12 on: Jan 4th, 2010, 12:03pm » |
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Yes, I have. Do you have something better in mind?
Then, I asked myself something else:
So far we have looked into the last 3-digits being greater than the first 3-digits. How about the first 3-digits being greater than the last 3-digits
E.g. a 6-digit number that could look like: ab4ab3
Then, i could write it as 1001y + 1000 = x^2
This presents a problem since 1000 is not a square... but i could find squares with 3 digits or more.
If this is becoming too annoying because i'm spending too much time on this, then let me know.
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ThudnBlunder
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Re: a 6-digit number
« Reply #13 on: Jan 4th, 2010, 1:49pm » |
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on Jan 4th, 2010, 12:03pm, BenVitale wrote:Yes, I have. Do you have something better in mind?
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Yes, you could solve the linear equations I gave and check if they have any solutions in range (using 2a rather than 2).
Better still, you can just plug the coefficients into an online solver.
For example, for a = 2 we obtain
7k = 143m 4 (no solutions in range)
11k = 91m 4 (k = 50, m = 6 gives x = 548, y = 300)
13k = 77m 4 (k = 53, m = 9 gives x = 691, y = 477)
It was because you were missing some solutions that I suspected you were using laborious trial and error.
on Jan 4th, 2010, 12:03pm, BenVitale wrote:
Then, i could write it as 1001y + 1000 = x^2
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Then x2 = 1001n - 1 (where n = y + 1).
or
n = (x2 + 1)/7*11*13
Unlike before, this poses a bit of a problem as x2 + 1 does not factorize (which does not mean there are no solutions). 
on Jan 4th, 2010, 12:03pm, BenVitale wrote:
If this is becoming too annoying because i'm spending too much time on this, then let me know.
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No, no, not at all. 
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| « Last Edit: Jan 7th, 2010, 5:17am by ThudnBlunder » |
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Benny
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Re: a 6-digit number
« Reply #14 on: Jan 6th, 2010, 4:02pm » |
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Thank you very much. It's clear, now.
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Benny
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Re: a 6-digit number
« Reply #15 on: Jan 7th, 2010, 1:50pm » |
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I'm curious about 6-digit numbers of the form xyzxyz (= 1001 * N)
Suppose that a number P = xyzxyz represents the area of a primitive Pythagorean triangle.
So, the sides of that primitive Pythagorean triangle are:
m2 - n2, 2mn, m2 + n2
The area A = mn(m - n)(m + n)
I went and checked on this site http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html
to see if I can find ways to generate m and n, then to be able to find other primitive Pythagorean triangles whose areas are of this form xyzxyz
I didn't find what I was looking for on that link.
Your thoughts, please.
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ThudnBlunder
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Re: a 6-digit number
« Reply #16 on: Jan 8th, 2010, 7:15am » |
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on Jan 7th, 2010, 1:50pm, BenVitale wrote:
Suppose that a number P = xyzxyz represents the area of a primitive Pythagorean triangle.
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Using my previous method I find that the values of m that work are too large to give a y < 1000
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pex
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Re: a 6-digit number
« Reply #17 on: Jan 8th, 2010, 7:34am » |
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on Jan 8th, 2010, 7:15am, ThudanBlunder wrote:
Using my previous method I find that the values of m that work are too large to give a y < 1000
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Strange: a simple numerical search reveals 24 solutions. I like the triple with sides (693, 1924, 2045), generated by (m, n) = (37, 26), best. 
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