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

From henhanna@gmail.com@21:1/5 to All on Wed Jun 29 11:25:35 2022
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.

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• From Stephen Perry@21:1/5 to henh...@gmail.com on Wed Jun 29 14:52:31 2022
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

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• From henhanna@gmail.com@21:1/5 to stephen...@gmail.com on Wed Jun 29 15:04:19 2022
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 ?

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• From Richard Heathfield@21:1/5 to Stephen Perry on Thu Jun 30 02:57:45 2022
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

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• From Edward Murphy@21:1/5 to Stephen Perry on Wed Jun 29 21:09:26 2022
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.

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• From Richard Heathfield@21:1/5 to Edward Murphy on Thu Jun 30 05:55:21 2022
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]

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

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• From Richard Tobin@21:1/5 to emurphy42@zoho.com on Thu Jun 30 09:23:55 2022
In article <t9j7ln\$hel\$1@gioia.aioe.org>,
Edward Murphy <emurphy42@zoho.com> wrote:
[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.

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

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