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anonymous
 3 years ago
Determine all positive integers \(n\ge 3\) for which\[1+\dbinom{n}{1}+\dbinom{n}{2}+\dbinom{n}{3}\]is a divisor of \(2^{2000}\)
anonymous
 3 years ago
Determine all positive integers \(n\ge 3\) for which\[1+\dbinom{n}{1}+\dbinom{n}{2}+\dbinom{n}{3}\]is a divisor of \(2^{2000}\)

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experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1\[ \binom{n}{0} + \binom{n}{1} = \binom{n+1}{1}\] \[ \binom{n}{2} + \binom{n}{3} = \binom{n+1}{3}\] \[ n+1 + {(n + 1 )n (n1)\over 3} = {(n+1)(3 + n(n1))\over 3}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0isn't it 6 instead of 3

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1sorry ... forgot it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so\[{(n+1)(6 + n(n1))\over 6}=\frac{(n+1)(n^2n+6)}{6}\]is a divisor of \(2^{2000}\)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1where do you get this problems?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so it must be in the form\[\frac{(n+1)(n^2n+6)}{6}=2^m \ \ \ \ m\le2000\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0compititions,books,...

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1ah man ... i thought engineers don't like number theory.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1\[ (n+1)(n^2n+6) \times k=3 \times 2^{2000+1} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or we can state problem like this\[(n+1)(n^2n+6)=3\times2^{m+1} \ \ \ \ m\le2000\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1n = 2 is one solution

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1n = 3, 4, 5, 6 < these does not work

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1let me try some programming solution up to 100

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah that will give all solutions

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0guys can u find \[\gcd(n+1,n^2n+6)\]?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1regarding this problem ... i don't have approach ... matlab collapses after 2^100

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1probably i should recursively divide the quotient by 2 after dividing by 3

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1between 1 to 999 1 12 2 24 3 48 7 384 23 12288

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0exper those are only solutions gane 1? how

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1i started loop from 1 clc; for i=1:999 num = (i+1)*(i^2  i + 6); if rem(num, 3) == 0 quo = num/3; while true if quo == 2; disp([i, num]); break; end; if rem(quo, 2) ~= 0; break; end; quo = quo/2; end end end

ganeshie8
 3 years ago
Best ResponseYou've already chosen the best response.0i tried euclid its repeating forever

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1probably are the only solutions (3, 48), (7, 384), (23, 12288)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0emm... there is useful formula\[\gcd(a,b)=\gcd(a,ak+b) \ \ \ k \in \mathbb{Z}\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1\[ \gcd(n+1, n(n+1)  2n + 6) => \gcd(n+1, 2(n 3))\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\gcd(n+1,8)= \gcd(n+1, 2(n+1)+2(n 3))\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1let's try elimination here ... this is divisible by 2 all times \[ (n+1)(n^2n+6)\] suppose what must be the value of n^2  n + 6 if n+1 is not divisible by 3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the point is \[\gcd(n+1,n^2n+6)8\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1n+1 = 3k +1, 3k +2 => n = 3k, 3k+1 9k^2  3k+6 < divisible by 3 9k^2 + 6k +1  3k  1 + 6 < divisible by 3 < bad luck

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1this seems to be valid for 3k, 3k+1, 3k+2 let's try for this 3k+2 (3k+3)(9k^2 + 12k + 4  3k  2 + 6) = 3(k+1)(9k^2 + 9 k + 4) (k+1)(9k^2 + 9 k + 4) = 2^x

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1implies k should be odd ... (2y+1) .. such that k+1 = 2^p for some p ie, k+1 = 2, 4, 8, 16, ....

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1for k = 7, the expression (9k^2 + 9 k + 4) is 512 ... which shows that n = 2*7 + 2 = 23

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1it remains to prove that there is no other n for the this type of 3k+2 form ... let k= 2^p  1 (9(2^p 1) ^2 + 9 (2^9 1) + 4) = 9*2^(2p)  18 2^p + 9 + 9 * 2^p  9 + 4 9*2^(2p)  9 *2^p + 4 = 2^q for some integer.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1man this is tough .. gotta work on classical + statistical mechanics ... have exam five days later.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0np man...how many exams do u have?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.11 down three more to go ... + 2 practicals

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh man go back to ur work :)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1yeah ... i'll watch the developments. physicist are not so good with numbers.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we have lots of times to work on brainwashing problems like this

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I wrote a Mathematica Program for \( n \le10^6 \) I found that \[ \frac{1}{6} (n+1) \left(n^2n+6\right) \] is a power of 2 for for \[ n=3, \quad \frac{1}{6} (n+1) \left(n^2n+6\right)=8\\ n=7, \quad \frac{1}{6} (n+1) \left(n^2n+6\right)=64\\ n=23, \quad \frac{1}{6} (n+1) \left(n^2n+6\right)=2048\\ \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I just did it \(n = 2 (10^6)\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here is the Mathematica Program Clear[h, n] h[n_] = 1/6 (1 + n) (6  n + n^2); PowerTwoQ[n_] := Module[{x }, x = n; While[ Divisible[x, 2] && x > 1, x = x/2;]; Return[ x == 1]] Q = {}; For [ n = 2, n < 2000000, n++; If [PowerTwoQ[h[n]], Q = Append[Q, {n, h[n]}]]]; Q The output is {{3, 8}, {7, 64}, {23, 2048}}

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[1+\dbinom{n}{1}+\dbinom{n}{2}+\dbinom{n}{3}=\frac{n^3+5n+6}{6}\] so we must have\[\frac{n^3+5n+6}{6}=2^k \ \ \ \ \ \ \ \ k\le2000\]or\[(n+1)(n^2n+6)=3\times2^{k+1}\]but notice that\[\gcd(n+1\ , \ n^2n+6)=\gcd(n+1\ , \ n^2nn^2n+6)\]\[\gcd(n+1\ , \ 2n+6)=\gcd(n+1\ , \ 2n+2n+2+6)=\gcd(n+1 \ , \ 8)\]\[\Rightarrow \gcd(n+1\ , \ n^2n+6)  8\]since \(n^2n+6>n+1\) power of \(2\) in \(n+1\) can not be greater than \(4\) in other words \(n+1=16\) or \(n+1 3\times 2^3\) so we have few cases to check\[n=3,5,7,11,15,23\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0checking that numbers gives \(3\) solution : \(\color\red{n=3,7,23}\)
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