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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 \(2^{2000}\)

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\[ \binom{n}{0} + \binom{n}{1} = \binom{n+1}{1}\] \[ \binom{n}{2} + \binom{n}{3} = \binom{n+1}{3}\] \[ n+1 + {(n + 1 )n (n-1)\over 3} = {(n+1)(3 + n(n-1))\over 3}\]
isn't it 6 instead of 3
sorry ... forgot it

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so\[{(n+1)(6 + n(n-1))\over 6}=\frac{(n+1)(n^2-n+6)}{6}\]is a divisor of \(2^{2000}\)
where do you get this problems?
so it must be in the form\[\frac{(n+1)(n^2-n+6)}{6}=2^m \ \ \ \ m\le2000\]
compititions,books,...
ah man ... i thought engineers don't like number theory.
lol
\[ (n+1)(n^2-n+6) \times k=3 \times 2^{2000+1} \]
or we can state problem like this\[(n+1)(n^2-n+6)=3\times2^{m+1} \ \ \ \ m\le2000\]
n = 2 is one solution
n = 3, 4, 5, 6 <-- these does not work
let me try some programming solution up to 100
yeah that will give all solutions
guys can u find \[\gcd(n+1,n^2-n+6)\]?
regarding this problem ... i don't have approach ... matlab collapses after 2^100
probably i should recursively divide the quotient by 2 after dividing by 3
between 1 to 999 1 12 2 24 3 48 7 384 23 12288
gcd is 1
exper those are only solutions gane 1? how
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
i tried euclid its repeating forever
probably are the only solutions (3, 48), (7, 384), (23, 12288)
emm... there is useful formula\[\gcd(a,b)=\gcd(a,ak+b) \ \ \ k \in \mathbb{Z}\]
\[ \gcd(n+1, n(n+1) - 2n + 6) => \gcd(n+1, 2(n -3))\]
\[\gcd(n+1,8)= \gcd(n+1, -2(n+1)+2(n -3))\]
let's try elimination here ... this is divisible by 2 all times \[ (n+1)(n^2-n+6)\] suppose what must be the value of n^2 - n + 6 if n+1 is not divisible by 3
the point is \[\gcd(n+1,n^2-n+6)|8\]
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
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
implies k should be odd ... (2y+1) .. such that k+1 = 2^p for some p ie, k+1 = 2, 4, 8, 16, ....
for k = 7, the expression (9k^2 + 9 k + 4) is 512 ... which shows that n = 2*7 + 2 = 23
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.
man this is tough .. gotta work on classical + statistical mechanics ... have exam five days later.
np man...how many exams do u have?
1 down three more to go ... + 2 practicals
oh man go back to ur work :)
yeah ... i'll watch the developments. physicist are not so good with numbers.
we have lots of times to work on brainwashing problems like this
sure
I wrote a Mathematica Program for \( n \le10^6 \) I found that \[ \frac{1}{6} (n+1) \left(n^2-n+6\right) \] is a power of 2 for for \[ n=3, \quad \frac{1}{6} (n+1) \left(n^2-n+6\right)=8\\ n=7, \quad \frac{1}{6} (n+1) \left(n^2-n+6\right)=64\\ n=23, \quad \frac{1}{6} (n+1) \left(n^2-n+6\right)=2048\\ \]
I just did it \(n = 2 (10^6)\)
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}}
\[1+\dbinom{n}{1}+\dbinom{n}{2}+\dbinom{n}{3}=\frac{n^3+5n+6}{6}\] so we must have\[\frac{n^3+5n+6}{6}=2^k \ \ \ \ \ \ \ \ k\le2000\]or\[(n+1)(n^2-n+6)=3\times2^{k+1}\]but notice that\[\gcd(n+1\ , \ n^2-n+6)=\gcd(n+1\ , \ n^2-n-n^2-n+6)\]\[\gcd(n+1\ , \ -2n+6)=\gcd(n+1\ , \ -2n+2n+2+6)=\gcd(n+1 \ , \ 8)\]\[\Rightarrow \gcd(n+1\ , \ n^2-n+6) | 8\]since \(n^2-n+6>n+1\) 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\[n=3,5,7,11,15,23\]
checking that numbers gives \(3\) solution : \(\color\red{n=3,7,23}\)

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