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interests / rec.puzzles / Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) are divisible by 10000 ?

SubjectAuthor
* for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) arehenh...@gmail.com
`* Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) areStephen Perry
 +- Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) arehenh...@gmail.com
 +- Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) areRichard Heathfield
 `* Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) areEdward Murphy
  +- Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) areRichard Heathfield
  `- Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) areRichard Tobin

1
for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) are divisible by 10000 ?

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Subject: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) are
divisible by 10000 ?
From: henha...@gmail.com (henh...@gmail.com)
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 by: henh...@gmail.com - Wed, 29 Jun 2022 18:25 UTC

a is an odd integer ---------- ( 0 < a < 10000)

Which values of ( a^2 - a ) are divisible by 10000 ?

pls wait a few days before posting answers or hints.

Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) are divisible by 10000 ?

<555132b8-a6af-440f-828c-480bb7800debn@googlegroups.com>

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Subject: Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) are
divisible by 10000 ?
From: stephen....@gmail.com (Stephen Perry)
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 by: Stephen Perry - Wed, 29 Jun 2022 21:52 UTC

On Wednesday, June 29, 2022 at 2:25:39 PM UTC-4, henh...@gmail.com wrote:
> a is an odd integer ---------- ( 0 < a < 10000)
>
> Which values of ( a^2 - a ) are divisible by 10000 ?
>
>
>
>
> pls wait a few days before posting answers or hints.

1, of course, is obvious. the other, not so much.

swp

Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) are divisible by 10000 ?

<a5b05464-ab03-4dff-8a6d-bcf5b30a93b5n@googlegroups.com>

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Subject: Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) are
divisible by 10000 ?
From: henha...@gmail.com (henh...@gmail.com)
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 by: henh...@gmail.com - Wed, 29 Jun 2022 22:04 UTC

On Wednesday, June 29, 2022 at 2:52:34 PM UTC-7, stephen...@gmail.com wrote:
> On Wednesday, June 29, 2022 at 2:25:39 PM UTC-4, henh...@gmail.com wrote:
> > a is an odd integer ---------- ( 0 < a < 10000)
> >
> > Which values of ( a^2 - a ) are divisible by 10000 ?
> >
> >
> >
> >
> > pls wait a few days before posting answers or hints.

> 1, of course, is obvious. the other, not so much.
>
> swp

a slightly better problem, without the ODD limitation.

is there another good variant ?

for integer a ( 0 < a < 100) Which ( a^2 ) are divisible by 100 ?

for integer a ( 0 < a < 100) Which ( a^2 - 1 ) are divisible by 100 ?

for integer a ( 0 < a < 100) Which ( a^2 +a -1 ) are divisible by 10 ?

Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) are divisible by 10000 ?

<t9ivup$1m7oo$1@dont-email.me>

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From: rjh...@cpax.org.uk (Richard Heathfield)
Newsgroups: rec.puzzles
Subject: Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) are
divisible by 10000 ?
Date: Thu, 30 Jun 2022 02:57:45 +0100
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 by: Richard Heathfield - Thu, 30 Jun 2022 01:57 UTC

On 29/06/2022 10:52 pm, Stephen Perry wrote:
> On Wednesday, June 29, 2022 at 2:25:39 PM UTC-4, henh...@gmail.com wrote:
>> a is an odd integer ---------- ( 0 < a < 10000)
>>
>> Which values of ( a^2 - a ) are divisible by 10000 ?
>>
>>
>>
>>
>> pls wait a few days before posting answers or hints.
>
> 1, of course, is obvious. the other, not so much.

Nor is the other other.

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) are divisible by 10000 ?

<t9j7ln$hel$1@gioia.aioe.org>

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From: emurph...@zoho.com (Edward Murphy)
Newsgroups: rec.puzzles
Subject: Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) are
divisible by 10000 ?
Date: Wed, 29 Jun 2022 21:09:26 -0700
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 by: Edward Murphy - Thu, 30 Jun 2022 04:09 UTC

On 6/29/2022 2:52 PM, Stephen Perry wrote:

> On Wednesday, June 29, 2022 at 2:25:39 PM UTC-4, henh...@gmail.com wrote:
>> a is an odd integer ---------- ( 0 < a < 10000)
>>
>> Which values of ( a^2 - a ) are divisible by 10000 ?
>>
>>
>>
>>
>> pls wait a few days before posting answers or hints.
>
> 1, of course, is obvious. the other, not so much.

[spoiler space]

a = 2b + 1 for some integer b (0 <= b <= 4999)

a^2 - a = (a - 1) * a
= 2b * (2b + 1)

If this is divisible by 10000, then
b * (2b + 1)
= b * a
is divisible by 5000 = 2^3 * 5^4.

2b + 1 can't be divisible by 2, so b must provide all three factors of
2, i.e. it must be divisible by 2^3, i.e. b = 8c and a = 16c + 1 for
some integer c (0 <= c <= 624).

b and 2b + 1 can't both be divisible by 5, so either a or b must
provide all four factors of 5.

* If b provides them, then the smallest possible values are
b = 0, a = 1, valid solution
followed by
b = 5000, a = 10001, too large

* If a provides them, then it's a multiple of 5^4, and odd (because
it also equals 2b + 1). This starts out with:
a = 625, b = 312 = 8 * 39, valid solution
Now note that adding 1250 to a is equivalent to adding 625 to b, and
625 = 8 * 78 + 1, so we need to do it 8 times before b is once again
divisible by 8:
a = 10625, b = 5312, too large

So a = 1 and a = 625 are the only solutions within the desired range.

Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) are divisible by 10000 ?

<t9jabp$1n4ef$1@dont-email.me>

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From: rjh...@cpax.org.uk (Richard Heathfield)
Newsgroups: rec.puzzles
Subject: Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) are
divisible by 10000 ?
Date: Thu, 30 Jun 2022 05:55:21 +0100
Organization: Fix this later
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 by: Richard Heathfield - Thu, 30 Jun 2022 04:55 UTC

On 30/06/2022 5:09 am, Edward Murphy wrote:
> On 6/29/2022 2:52 PM, Stephen Perry wrote:
>
>> On Wednesday, June 29, 2022 at 2:25:39 PM UTC-4,
>> henh...@gmail.com wrote:
>>> a is an odd integer ---------- ( 0 < a < 10000)
>>>
>>> Which values of ( a^2 - a ) are divisible by 10000 ?
>>>
>>>
>>>
>>>
>>> pls wait a few days before posting answers or hints.
>>
>> 1, of course, is obvious.  the other, not so much.
>
> [spoiler space]
>
>
>
>
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>
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>
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> a = 2b + 1 for some integer b (0 <= b <= 4999)
>
> a^2 - a = (a - 1) * a
>         = 2b * (2b + 1)
>
> If this is divisible by 10000, then
>           b * (2b + 1)
>         = b * a
> is divisible by 5000 = 2^3 * 5^4.
>
> 2b + 1 can't be divisible by 2, so b must provide all three
> factors of
> 2, i.e. it must be divisible by 2^3, i.e. b = 8c and a = 16c + 1 for
> some integer c (0 <= c <= 624).
>
> b and 2b + 1 can't both be divisible by 5, so either a or b must
> provide all four factors of 5.
>
>   * If b provides them, then the smallest possible values are
>       b = 0, a = 1, valid solution
>     followed by
>       b = 5000, a = 10001, too large
>
>   * If a provides them, then it's a multiple of 5^4, and odd
> (because
>     it also equals 2b + 1). This starts out with:
>       a = 625, b = 312 = 8 * 39, valid solution
>     Now note that adding 1250 to a is equivalent to adding 625
> to b, and
>     625 = 8 * 78 + 1, so we need to do it 8 times before b is
> once again
>     divisible by 8:
>       a = 10625, b = 5312, too large
>
> So a = 1 and a = 625 are the only solutions within the desired
> range.

Aha! An ODD integer is required. I missed that, giving me a
four-digit solution.

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) are divisible by 10000 ?

<t9jq3b$2ot$2@macpro.inf.ed.ac.uk>

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From: rich...@cogsci.ed.ac.uk (Richard Tobin)
Newsgroups: rec.puzzles
Subject: Re: for odd a ( 0 < a < 10000) Which values of ( a^2 - a ) are
divisible by 10000 ?
Date: Thu, 30 Jun 2022 09:23:55 +0000 (UTC)
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 by: Richard Tobin - Thu, 30 Jun 2022 09:23 UTC

In article <t9j7ln$hel$1@gioia.aioe.org>,
Edward Murphy <emurphy42@zoho.com> wrote:
>[spoiler space]
>
>
>
>
>
>
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>
>
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>a = 2b + 1 for some integer b (0 <= b <= 4999)
>
>a^2 - a = (a - 1) * a
> = 2b * (2b + 1)
>
>If this is divisible by 10000, then
> b * (2b + 1)
> = b * a
>is divisible by 5000 = 2^3 * 5^4.
>
>2b + 1 can't be divisible by 2, so b must provide all three factors of
>2, i.e. it must be divisible by 2^3, i.e. b = 8c and a = 16c + 1 for
>some integer c (0 <= c <= 624).
>
>b and 2b + 1 can't both be divisible by 5, so either a or b must
>provide all four factors of 5.
>
> * If b provides them, then the smallest possible values are
> b = 0, a = 1, valid solution
> followed by
> b = 5000, a = 10001, too large
>
> * If a provides them, then it's a multiple of 5^4, and odd (because
> it also equals 2b + 1). This starts out with:
> a = 625, b = 312 = 8 * 39, valid solution
> Now note that adding 1250 to a is equivalent to adding 625 to b, and
> 625 = 8 * 78 + 1, so we need to do it 8 times before b is once again
> divisible by 8:
> a = 10625, b = 5312, too large
>
>So a = 1 and a = 625 are the only solutions within the desired range.

To put it another way: either a-1 must be a multiple of 10000 or
a must be a multiple of 5^4 that is equal to 1 mod 2^4 (=16). 625 is
equal to 1 mod 16, so it's a solution; further solutions will be of
the form 625 (16k+1).

That is, the solutions are

10000 k + 1
and
10000 k + 625

-- Richard

1
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