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.
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
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.
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.
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.
[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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