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

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

mukushlaBest ResponseYou've already chosen the best response.1
isn't it 6 instead of 3
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
sorry ... forgot it
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.1
where do you get this problems?
 one year ago

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

mukushlaBest ResponseYou've already chosen the best response.1
compititions,books,...
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
ah man ... i thought engineers don't like number theory.
 one year ago

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

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

experimentXBest ResponseYou've already chosen the best response.1
n = 2 is one solution
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
n = 3, 4, 5, 6 < these does not work
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
let me try some programming solution up to 100
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
yeah that will give all solutions
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
guys can u find \[\gcd(n+1,n^2n+6)\]?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
regarding this problem ... i don't have approach ... matlab collapses after 2^100
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
probably i should recursively divide the quotient by 2 after dividing by 3
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
between 1 to 999 1 12 2 24 3 48 7 384 23 12288
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
exper those are only solutions gane 1? how
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
i 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
 one year ago

ganeshie8Best ResponseYou've already chosen the best response.0
i tried euclid its repeating forever
 one year ago

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

mukushlaBest ResponseYou've already chosen the best response.1
emm... there is useful formula\[\gcd(a,b)=\gcd(a,ak+b) \ \ \ k \in \mathbb{Z}\]
 one year ago

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

mukushlaBest ResponseYou've already chosen the best response.1
\[\gcd(n+1,8)= \gcd(n+1, 2(n+1)+2(n 3))\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
let'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
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
the point is \[\gcd(n+1,n^2n+6)8\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
n+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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
this 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
 one year ago

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

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

experimentXBest ResponseYou've already chosen the best response.1
it 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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
man this is tough .. gotta work on classical + statistical mechanics ... have exam five days later.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
np man...how many exams do u have?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
1 down three more to go ... + 2 practicals
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
oh man go back to ur work :)
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yeah ... i'll watch the developments. physicist are not so good with numbers.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
we have lots of times to work on brainwashing problems like this
 one year ago

eliassaabBest ResponseYou've already chosen the best response.0
I 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 \[ n=3, \quad \frac{1}{6} (n+1) \left(n^2n+6\right)=8\\ n=7, \quad \frac{1}{6} (n+1) \left(n^2n+6\right)=64\\ n=23, \quad \frac{1}{6} (n+1) \left(n^2n+6\right)=2048\\ \]
 one year ago

eliassaabBest ResponseYou've already chosen the best response.0
I just did it \(n = 2 (10^6)\)
 one year ago

eliassaabBest ResponseYou've already chosen the best response.0
Here 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}}
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
\[1+\dbinom{n}{1}+\dbinom{n}{2}+\dbinom{n}{3}=\frac{n^3+5n+6}{6}\] 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\[n=3,5,7,11,15,23\]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
checking that numbers gives \(3\) solution : \(\color\red{n=3,7,23}\)
 one year ago
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