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 3 years ago
Determine all positive integers \(n\ge 3\) such that \(2^{2000}\) is divisible by
\(1+\left(\begin{matrix}n \\ 1\end{matrix}\right)+\left(\begin{matrix}n\\ 2\end{matrix}\right)+\left(\begin{matrix}n\\ 3\end{matrix}\right)\).
 3 years ago
Determine all positive integers \(n\ge 3\) such that \(2^{2000}\) is divisible by \(1+\left(\begin{matrix}n \\ 1\end{matrix}\right)+\left(\begin{matrix}n\\ 2\end{matrix}\right)+\left(\begin{matrix}n\\ 3\end{matrix}\right)\).

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SqueeSpleen
 3 years ago
Best ResponseYou've already chosen the best response.1Combinatory numbers?

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.3Latex tip for Mr.Math : \(\binom{n}{r} \)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1or even \[\dbinom{n}{k}\]

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.3OR even \( n \choose r \)

SqueeSpleen
 3 years ago
Best ResponseYou've already chosen the best response.1Impair = odd number, I searched the correct translation xD, I don't know how to write in English the numbers of the kind of 2k where k is an integer.

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.0I don't understand, what's the difference between mine and yours? @Fool and satellite

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.3Your need more keystrokes :P

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.0Oh I got it. Thanks :)

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.0@SqueeSpleen: If you could write it in latex, that would be great!

SqueeSpleen
 3 years ago
Best ResponseYou've already chosen the best response.1I made some mistakes, I'll start from zero, and I may use latex.

SqueeSpleen
 3 years ago
Best ResponseYou've already chosen the best response.1\[1+\left(\begin{matrix}n \\ 1\end{matrix}\right)+\left(\begin{matrix}n \\ 2\end{matrix}\right)+\left(\begin{matrix}n \\ 3\end{matrix}\right)=1+n+n(n+1)/2+n(n+1)(n+2)/6\] \[=1+n(1+(n+1)/2+(n+1)(n+2)/6)=1+n(1+((n+1)/2)(1+(n+2)/3)\] Or \[1+n!/((n1)!1!)+n!/((n2)!2!)+n!/(n3!3!)=\] \[=1+n+n*(n1)/2+n*(n1)*(n2)/6=\] \[=(6+6n+3n^23n+n^3+2n3n^2)/6=\] \[=(n^3+5n+6)/6\] \[=n(n^2+5)/6+1\] I used the division by 6 as if 6 were a prime number, so it was a mistake. n by 2 or 3, or n^2+5 has to be divided by 3 or 2. Let's try: n=6k It's key. n=6k+1 then: \[n(n^2+5)/6+1=36 k^3+18 k^2+8 k+1\] But then you get and odd number, you can't divide it by 6. So n isn't of the kind of 6k+1 It's n=6k+2 then \[n(n^2+5)/6+1=36 k^3+36 k^2+17 k+3\] It can be divided by 3 only if k is multiple of 3 and cannot be divided by 2 n=6k+3 \[n(n^2+5)/6+1=36 k^3+54 k^2+32 k+7\] Then I can't be divided by 2, then it can't be divided by 2. \[n=6k'+4=6k2\] \[n(n^2+5)/6+1=36 k^336 k^2+17 k3\] It can be divided by 3 only if k is multiple of 3 and cannot be divided by 2 n=6k'+5=6k1 \[n(n^2+5)/6+1=36 k^318 k^2+8 k1\] But then you get and odd number, you can't divide it by 6. This means the number is of the kind of: 6k, 6k+1, 6k+3 or 6k+5 Where k is an integer. I barely eliminated some choices, so I'll try to take other way to solve this.

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0Shouldn't it be \( \dbinom{n}{2} = \large\frac{n(n1)}{2} \)

SqueeSpleen
 3 years ago
Best ResponseYou've already chosen the best response.1Yes I made the mistake in the first lines, but I didn't in the following lines, thanks for the advice, I don't know how to edit messages.

SqueeSpleen
 3 years ago
Best ResponseYou've already chosen the best response.1I found n=3, n=7 and n=23 but I can't prove that they're the only solutions. I think they're but I have not idea how to prove this.

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0I like your dedication! 17 hours and you still wanna solve it! excellent!

SqueeSpleen
 3 years ago
Best ResponseYou've already chosen the best response.1I was trying to prove that if you pick k big enough then n(n^2+5)/6+1 will not be a power of 2, but I got nothing except some less paper and ink :P

mukushla
 2 years ago
Best 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\[n^3+5n+6=3\times2^{k+1}\]\[(n+1)(n^2n+6)=3\times2^{k+1}\]note 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 \(n^2n+6>n+1\) so its immidiate that\[\]\(n+1=16\) or \(n+13\times 2^3\) so we have few cases to check : \(n=1,2,3,5,7,11,15,23\)
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