Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- Mr.Math

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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- Mr.Math

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

Combinatory numbers?

- anonymous

Latex tip for Mr.Math : \(\binom{n}{r} \)

- anonymous

or even
\[\dbinom{n}{k}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

OR even \( n \choose r \)

- anonymous

Impair = 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

I don't understand, what's the difference between mine and yours? @Fool and satellite

- anonymous

Your need more keystrokes :P

- Mr.Math

Oh I got it. Thanks :-)

- Mr.Math

\[n\choose r\]

- Mr.Math

@SqueeSpleen: If you could write it in latex, that would be great!

- anonymous

I made some mistakes, I'll start from zero, and I may use latex.

- Mr.Math

Okay.

- anonymous

\[1+\left(\begin{matrix}n \\ 1\end{matrix}\right)+\left(\begin{matrix}n \\ 2\end{matrix}\right)+\left(\begin{matrix}n \\ 3\end{matrix}\right)=1+n+n(n+1)/2+n(n+1)(n+2)/6\]
\[=1+n(1+(n+1)/2+(n+1)(n+2)/6)=1+n(1+((n+1)/2)(1+(n+2)/3)\]
Or
\[1+n!/((n-1)!1!)+n!/((n-2)!2!)+n!/(n-3!3!)=\]
\[=1+n+n*(n-1)/2+n*(n-1)*(n-2)/6=\]
\[=(6+6n+3n^2-3n+n^3+2n-3n^2)/6=\]
\[=(n^3+5n+6)/6\]
\[=n(n^2+5)/6+1\]
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.
\[n=6k'+4=6k-2\]
\[n(n^2+5)/6+1=36 k^3-36 k^2+17 k-3\]
It can be divided by 3 only if k is multiple of 3 and cannot be divided by 2
n=6k'+5=6k-1
\[n(n^2+5)/6+1=36 k^3-18 k^2+8 k-1\]
But then you get and odd number, you can't divide it by 6.
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.

- anonymous

Shouldn't it be \( \dbinom{n}{2} = \large\frac{n(n-1)}{2} \)

- anonymous

Yes 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.

- anonymous

I 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.

- anonymous

I like your dedication! 17 hours and you still wanna solve it! excellent!

- anonymous

I 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

- anonymous

\[1+\dbinom{n}{1}+\dbinom{n}{2}+\dbinom{n}{3}=\frac{n^3+5n+6}{6}\]so we must have\[\frac{n^3+5n+6}{6}=2^k \ \ \ \ \ \ \ \ k\le2000 \]or\[n^3+5n+6=3\times2^{k+1}\]\[(n+1)(n^2-n+6)=3\times2^{k+1}\]note 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 \(n^2-n+6>n+1\) so its immidiate that\[\]\(n+1=16\) or \(n+1|3\times 2^3\) so we have few cases to check :
\(n=1,2,3,5,7,11,15,23\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.