Anonymous at Fri, 13 Dec 2024 20:43:04 UTC No. 16511929

>>16511903
000000000000000000000000000000

Anonymous at Fri, 13 Dec 2024 20:43:59 UTC No. 16511932

>>16511903
21i^2 >= 10^29

i >= sqrt(10^29/21)

37j^2 >= 10^29 - 21i^2

j>= sqrt((10^29 - 21i^2)/37)

Anonymous at Sat, 14 Dec 2024 02:57:40 UTC No. 16512266

>>16511903
all you need is two for loops to check all (x,y) with 0 < x < 10^15 and 0 < y < 10^15 to find the nearest one with value at least 10^30 and you are done.

Anonymous at Sat, 14 Dec 2024 03:05:12 UTC No. 16512272

>>16512266
You don't need two loops. If you have one fixed y then the optimal x for it is
[eqn]x = \text{ceil} \left( \sqrt{\frac{10^{30} - 37y^2}{21}} \right) [/eqn]
But even then checking 10^15 cases would take days of computing. You have to find a smarter method.

Anonymous at Sat, 14 Dec 2024 03:07:53 UTC No. 16512274

>>16512272
>Be smarter
It doesn't work like that

Anonymous at Sat, 14 Dec 2024 04:17:48 UTC No. 16512323

>>16511903
Smallest 30 digit number is 10^29. Take the squareroot of 10^29 and round it up (A) and divide it separately by 21 and 37 to two numbers that you then round down (B and C respectively). Create two loops, one that goes from B to A and another from B to C. One loop solves for y and has x = 0 and another solves for x and has y set to 0. Whichever loop provides the smallest 30 digit number will give the answer. Should take less than a second to solve with a modern computer.

Anonymous at Sat, 14 Dec 2024 04:44:21 UTC No. 16512350

>>16512323
is this what you are suggesting?

Anonymous at Sat, 14 Dec 2024 05:03:36 UTC No. 16512359

>>16511903
You guys are all so low IQ. Just solve the x and y intercepts for (21x^2 + 37y^2 ≥ 10^29) then round the y and x intercepts up to the next integer.

Then solve for the answer.

Anonymous at Sat, 14 Dec 2024 05:13:26 UTC No. 16512362

>>16512359
So
21x^2 = 10^29
So
x = 69006555934235.4
Call it 69006555934236

And

37y^2 = 10^29
So
y=51987524491003.6
Call it 51987524491004

So answer is
21((69006555934236)^2) + 37((51987524491004)^2)

Which is

200000000000003082085141314208

SOLVED

Anonymous at Sat, 14 Dec 2024 05:15:41 UTC No. 16512363

>>16512359
>>16512362
The 10 minutes break between my answers was me having a shower (it's 5am here in the UK).

Anonymous at Sat, 14 Dec 2024 05:20:40 UTC No. 16512365

>>16512359
>>16512362
This gives 0 guarantees of it being the smallest 30 digit number available

Anonymous at Sat, 14 Dec 2024 05:28:41 UTC No. 16512368

>>16512362
try for 16 digit case and try beat x = 4857985, y = 3692211

Anonymous at Sat, 14 Dec 2024 05:36:56 UTC No. 16512372

>>16512365
You're right! I forgot to divide by 2!

21x^2 = 0.5 * 10^29

x = 48795003647426.7
Call it 48795003647427

37y^2 = 0.5 * 10^29

y = 36760731104690.4
Call it 36760731104691

New answer is

21((48795003647427)^2) + 37((36760731104691)^2)

So

100000000000002365308679729706

Apologies. It's early

Anonymous at Sat, 14 Dec 2024 05:46:00 UTC No. 16512375

>>16512365
Sure thing

21x^2 = 0.5 * 10^15
x = 4879500.36474267
Call it 4879501

37y^2 = 0.5 * 10^15
y = 3676073.11046904
Call it 3676073

21((4879501)^2) + 37((3676074^2))

1000000100138194

Anonymous at Sat, 14 Dec 2024 05:53:19 UTC No. 16512377

>>16512372
>21((48795003647427)^2) + 37((36760731104691)^2)

Also just realised I could have done

21((48795003647426)^2) + 37((36760731104691)^2)

Which gets you

100000000000000315918526537793

You can't go any smaller though

Feel free to try yourself:

https://www.wolframalpha.com/input?i=21%28%2848795003647426%29%5E2%29+%2B+37%28%2836760731104691%29%5E2%29

Anonymous at Sat, 14 Dec 2024 05:54:56 UTC No. 16512378

>>16512368
>>16512375
Oops wrong (you)

>>16512365
You're correct it's not 100% but it gets you within 1 digit basically.

See
>>16512377

Anonymous at Sat, 14 Dec 2024 05:57:21 UTC No. 16512381

>>16512378
21*4857985^2+37*3692211^2 = 1000000000000002 btw

Anonymous at Sat, 14 Dec 2024 06:06:24 UTC No. 16512385

>>16512381
Ignore everything I've said. I'm retarded.

>>16512359
>>16512362
>>16512363
>>16512372
>>16512375
>>16512377
>>16512378

What an idiot

Anonymous at Sat, 14 Dec 2024 06:26:47 UTC No. 16512402

>>16512381
WAIT! I've just realised

I should be calculating the closest point on an ellipse.

>>16512362

69006555934235.4 = a

51987524491003.6 = b

I can't do the maths right now.

But the closest point to the centre on the edge of the ellipse is the solution to the x and y

Anonymous at Sat, 14 Dec 2024 06:34:04 UTC No. 16512406

>>16512350
Yes

Anonymous at Sat, 14 Dec 2024 10:14:21 UTC No. 16512537

Coding it up. Now frustrated because something's wrong. Can't see what. Maybe this function?

FUNCTION FIND(BYVAL TARGET AS DOUBLE, _BASE AS DOUBLE, POWER AS DOUBLE) AS ULONGINT
IF TARGET <= 0 THEN RETURN 0
DIM Z AS DOUBLE
TARGET /= _BASE
Z = TARGET^(1 / POWER)
IF FIX(Z) <> Z THEN Z = FIX(Z) + 1
RETURN Z
END FUNCTION
Meant to find the smallest integer for x or y and some base and power that equals or exceeds a target value.

Rest of the code: (unimportant bits left out)

X_MAX = FIND(TARGET, X_BASE, X_POWER)
PRINT "SOLUTION NEEDS MAXIMUM OF X = " ; X_MAX

FOR X = 0 TO X_MAX
X_VAL = X_BASE * (X^X_POWER)
T_TARGET = TARGET - X_VAL
Y = FIND(T_TARGET, Y_BASE, Y_POWER)
T_VAL = X_VAL + Y_BASE * Y ^ Y_POWER
IF T_VAL < BEST_VAL OR BEST_VAL = 0 THEN
BEST_VAL = T_VAL
Y_BEST = Y
X_BEST = X
END IF
NEXT X

T_TARGET is a temporary thing, just the amount the y-term must be for a given x-term

Anonymous at Sat, 14 Dec 2024 16:56:02 UTC No. 16512806

>>16512266
>>16512272
Perhaps I should have picked a number larger than "thirty" to make clear that this type of solution is too slow. Feel free to swap "thirty" for a larger number.
For example for 50-digit number, the solution would be:
10000000000000000000000000000000000000000000000026 = 21x^2 + 37y^2, where
x=8466062101933517705547
y=519836118709208520290349

>>16512385
thank you for the attempt but it's not that simple
some knowledge or googling ability is required

Anonymous at Sat, 14 Dec 2024 17:17:30 UTC No. 16512832

>>16512806
>>16512385
>>16512402

Me again, just on the way back from work.

How about:

21x^2 + 37y^2 = 10^29

Making an ellipse

Now we can plot a line of intersection:

y = (21/37)x

Calculate positive points of intersection:

x = 100000000000000(sqrt(185/609))
y = 100000000000000(sqrt(105/1073))

These numbers would give you exactly 10^29

Rounding to integer values

x = 55115939898275
y = 31282019942264

This would give us
9.9999999999998998190364978877 * 10^28

(Wrong)

So we increase x by one digit:

x = 55115939898276
y = 31282019942264

100000000000001313059840706448

And then we do the same but y increased by one digit instead

x = 55115939898275
y = 31282019942265

100000000000001313059840706450


100000000000001313059840706448 (1st one) is smaller.

Therefore it should be

100000000000001313059840706448

With

x = 55115939898276
y = 31282019942264

I might be wrong again though but who knows.

I need to do the actual maths for the ellipse when I get home (calculate closest point to the origin)

But is that answer at least close?

Anonymous at Sat, 14 Dec 2024 18:14:11 UTC No. 16512876

>>16512832
[code]
55115939898275n 31282019942264n 99999999999998998190364978877n false
55115939898276n 31282019942264n 100000000000001313059840706448n true (This one)
55115939898275n 31282019942265n 100000000000001313059840706450n true
[/code]

Anonymous at Sat, 14 Dec 2024 18:27:18 UTC No. 16512883

>>16512806
>>16512832
Ok lol this is wrong again. Not having a good day. Home now though

Convert

21x^2 + 37y^2 to polar coordinates then

Calculate minimum for

r = (10^14 * sqrt(10*(21cos^2(θ)+37sin^2(θ))))/(21cos^2(θ)+37sin^2(θ))

Which gives us

x = 100000000000000(sqrt(10/37))

So

y = (400000000000000 sqrt(10))/37


Or rounded
x = 51987524491004
y = 34186785515334

21((51987524491004)^2) + 37((34186785515334)^2)

Gives

100000000000001226648933355908

Unless I've made a mistake again that is.

Anonymous at Sat, 14 Dec 2024 18:46:49 UTC No. 16512903

>>16512323
> less than a second to solve with a modern computer
Wrong, it need at least one day (all cores working at 100%) to a week if you don't want to toast the processor.

Anonymous at Sat, 14 Dec 2024 19:06:10 UTC No. 16512929

>>16512883
Lowest so far, but the real problem is to find a point of the ellipse that is as close as possible to the integer grid that results in a number of 30 digits. It could be anywhere.

Anonymous at Sat, 14 Dec 2024 19:07:10 UTC No. 16512933

>>16512929

Anonymous at Sat, 14 Dec 2024 23:33:51 UTC No. 16513237

>>16512929
True. There's got to be pattern to it though

>>16512806
>For example for 50-digit number, the solution would be:
>10000000000000000000000000000000000000000000000026 = 21x^2 + 37y^2, where
>x=8466062101933517705547
>y=519836118709208520290349

>>16512381
>21*4857985^2+37*3692211^2 = 1000000000000002 btw

Anonymous at Mon, 16 Dec 2024 04:18:43 UTC No. 16514354

100000000000000000000000000069