/sci/ - Science & Math » Thread #13422927

7KiB, 201x251, easter_pepe.jpg

View Same Google iqdb SauceNAO

Magic Trick

Kek Mage Thu 22 Jul 11:24:30 2021 No.13422927 View Reply Original Report

Quoted By: >>13422939 >>13423016 >>13425048 >>13425747

>step 1: pick a number, k, it can be any nonnegative integer.
>step 2: multiply that number by 12
>step 3: add 12 to the result of step 2
>step 4: raise 2 to the power of the result of step 3
>step 5: subtract 1 from the result of step 4

Your answer from step 5 will ALWAYS be divisible by 63. Have fun getting no sleep.

Anonymous

Anonymous Thu 22 Jul 2021 11:30:56 No.13422939 Report

Quoted By: >>13422942

>>13422927
1. okay, k=7
2. okay, 12*7= 84
3. 2^84
>2^84
Ain't doing that shit nigga.

Kek Mage

View Same Google iqdb SauceNAO pepe_bird2.jpg, 50KiB, 929x1175

Kek Mage Thu 22 Jul 2021 11:32:51 No.13422942 Report

Quoted By: >>13422952

>>13422939
You skipped step 3 like a tard. If you're too pussy, then pick a smaller number.

Anonymous

Anonymous Thu 22 Jul 2021 11:35:48 No.13422952 Report

Quoted By: >>13422956

>>13422942
>84+12
>96
>2^96
Thanks, it's even worse.

>then pick a smaller number.
It's a single digit number. How small do you fucking want?
I'm not multiplying two 96 times like an autist.

Kek Mage

Kek Mage Thu 22 Jul 2021 11:38:48 No.13422956 Report

Quoted By: >>13422958 >>13422965

>>13422952
Every modern computer worth a damn has 64-bit precision...

>>Here babby, have a cheat sheet...

>>2^12 = 63(65) + 1
>>2^24 = 63(266305) + 1
>>2^36 = 63(1090785345) + 1
>>2^48 = 63(4467856773185) + 1
>>2^60 = 63(18300341342965825) + 1
>>...

Kek Mage

Kek Mage Thu 22 Jul 2021 11:39:57 No.13422958 Report

Quoted By:

>>13422956
2^12 = 63(65) + 1
2^24 = 63(266305) + 1
2^36 = 63(1090785345) + 1
2^48 = 63(4467856773185) + 1
2^60 = 63(18300341342965825) + 1
...

fuckin greentext reeee

Anonymous

Anonymous Thu 22 Jul 2021 11:43:25 No.13422965 Report

Quoted By: >>13422971

>>13422956
64 Is smaller than 96 so I could not have done it even with. computer.

Kek Mage

Kek Mage Thu 22 Jul 2021 11:46:39 No.13422971 Report

Quoted By: >>13422973

>>13422965
2^64 - 1

range is [0, 2^64 - 1) for the largest unsigned integral type for any 64 bit processor.

Kek Mage

Kek Mage Thu 22 Jul 2021 11:47:41 No.13422973 Report

Quoted By:

>>13422971
[0,2^64-1]*
typo

Anonymous

View Same Google iqdb SauceNAO 1625822645057.jpg, 62KiB, 640x640

Anonymous Thu 22 Jul 2021 12:14:58 No.13423016 Report

Quoted By: >>13423033 >>13423051 >>13423153

>>13422927
2^(12k+ 12) - 1
= 2^(12(k + 1)) - 1
= 4096^(k + 1) - 1
= 4096*4096^k - 1

Note that 4096 = 63*65 + 1
Therefore, the above is equal to
(63*65 + 1)*4096^k - 1
= 63*65*4096^k + (4096^k - 1)
= 63*C1 + (4096^k - 1)

Now, if k >= 1, then you can basically repeat what we just did but with 4096^k - 1 instead of 4096^(k + 1) - 1.
I.e., the next step would be
63*C1 + (4096*4096^(k - 1) - 1)
= 63*C1 + (63*65 + 1)*4096^(k - 1) - 1
= 63*C1 + 63*65*4096^(k - 1) + (4096^(k - 1) - 1)
= 63*C1 + 63*C2 + (4096^(k - 1) - 1)

By induction, we will eventually reach k = 0, and the resulting expression will be:
63*C1 + ... + 63*Ck + (4096^0 - 1) = 63*(C1 + ... + Ck)

Thus, the originally proposed expression is always divisible by 63.

Kek Mage

View Same Google iqdb SauceNAO apu_browse3.jpg, 206KiB, 900x873

Kek Mage Thu 22 Jul 2021 12:23:28 No.13423033 Report

Quoted By:

>>13423016
party pooper...
at least someone has a brain here though.

Anonymous

Anonymous Thu 22 Jul 2021 12:35:42 No.13423051 Report

Quoted By: >>13423153

>>13423016
>anon is a nigger who can't into LaTeX
$ \begin{align*} 2^{12k + 12} - 1 &= 2^{12(k+1)} - 1\\ &= 4096^{k+1} - 1\\ &= 4096\cdot 4096^k - 1 \end{align*} $
Note that $4096 = 63\cdot 65 + 1$
Therefore, the above is equal to
$ \begin{align*} (63\cdot 65 + 1)\cdot 4096^k - 1 &=63\cdot 65\cdot 4096^k + (4096^k - 1)\\ &=63\cdot C_1 + (4096^k - 1) \end{align*} $ Now, if $k\geq 1$ , then you basically repeat what we just did but with $4096^k - 1$ instead of $4096^{k + 1} - 1$
I.e., the next step would be
$ \begin{align*} 63\cdot C_1 + (4096\cdot 4096^{k - 1} - 1) &= 63\cdot C_1 + (63\cdot 65 + 1)\cdot 4096^{k - 1} - 1\\ &= 63\cdot C_1 + 63\cdot 65\cdot 4096^{k - 1} + (4096^{k - 1} - 1)\\ &= 63\cdot C_1 + 63\cdot C_2 + (4096^{k - 1} - 1) \end{align*} $ By induction, we will eventually reach $k = 0$ and the resulting expression will be:
$63\cdot C_1 + \dots + 63\cdot C_k + (4096^0 - 1) = 63\cdot (C_1 + \dots + C_k)$
Thus, the originally proposed expression is always divisible by $63$ .
Good job however.

Anonymous

Anonymous Thu 22 Jul 2021 12:44:47 No.13423068 Report

Quoted By:

7 KB JPG
>step 1: pick a number, k, it can be any nonnegative integer.
>step 2: raise 10 to the power of the result of step 1
>step 3: subtract 1
Your number will always be divisible by 9. Have fun getting no sleep.

Anonymous

Anonymous Thu 22 Jul 2021 13:28:50 No.13423153 Report

Quoted By: >>13423214 >>13424516

>>13423016
>>13423051
Can (2^6 - 1), which is 63, just be factored out of 2^(12k + 12) - 1 more directly? Got some difference-of-squares shit happening there. I can't be bothered working it, though.

Anonymous

Anonymous Thu 22 Jul 2021 13:54:45 No.13423214 Report

Quoted By: >>13424516

>>13423153
>Got some difference-of-squares shit happening there
$ \begin{align*} 2^6 - 1 &= (2^3 - 1)(2^3 + 1)\\ &=(2 - 1)(2^2 + 2 + 1)(2 + 1)(2^2 - 2 + 1)\\ \end{align*} $
and
$ \begin{align*} 2^{12(k + 1)} - 1 &= (2^{6(k+1)} - 1)(2^{6(k+1)} + 1)\\ &= (2^{3(k+1)} - 1)(2^{3(k+1)} + 1)(2^{6(k+1)} + 1)\\ &=(2^{k+1} - 1)(2^{2(k+1)} + 2^{k+1} + 1)(2^{k+1} + 1)(2^{2(k+1)} - 2^{k+1} + 1)(2^{6(k+1)} + 1) \end{align*}\\ $
But I don't see how 63 can be factored out.

Anonymous

View Same Google iqdb SauceNAO E5TsM27XMAEPGHf.jpg, 93KiB, 1261x360

Anonymous Thu 22 Jul 2021 20:22:15 No.13424516 Report

Quoted By:

>>13423153
>>13423214
There is a "simpler" version of my proof but it would rely on the binomial theorem.
4096^(k + 1) - 1 = (63*65 + 1)^(k + 1) - 1

Using the binomial theorem on (63*65 + 1)^(k + 1), we will get a sum of all terms of the form Ca * (63*65)^a * 1^b, where 0 <= a,b <= k + 1 subject to the constraint a + b = k + 1.
Note that a > 0 implies that 63 will be a factor of the resulting term. Therefore, the entire expansion of 4096^(k + 1) can be re-written as:
63*K + (63*65)^0 *1^(k + 1) = 63*K + 1.

Therefore, (63*65 + 1)^(k + 1) - 1 = (63*K + 1) - 1 = 63*K. So this would be another way to prove that the original expression is divisible by 63.
However, I prefer elementary proofs (the original proof just requires induction) over ones that rely on previous results which is why I presented that one originally.

Anonymous

Anonymous Thu 22 Jul 2021 22:34:06 No.13425048 Report

Quoted By: >>13425226

>>13422927
>113
>113*12=1356
>1358
>1871424
>1871423/63=29705.1269841
?

Anonymous

View Same Google iqdb SauceNAO bonk.gif, 31KiB, 128x126

Anonymous Thu 22 Jul 2021 23:31:38 No.13425226 Report

Quoted By:

>>13425048
$ 113*12 = 1356\\ 1356+12 = 1368\\ 2^{1368} = 63n + 1 $

Anonymous

Anonymous Fri 23 Jul 2021 02:43:25 No.13425747 Report

Quoted By:

>>13422927
1. 1
2. 12
3. 24
4. 16777216
5. 16777215
16777215 / 63 = 266305
Wow OP, that's really cool!

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches