It is not locked? Participants may not read our messages? Thank you for rapidity, Valentin! For this problem, I think answer is only $1$ and it is clear, since for any prime $p>3$ we have $2^{p-2}+3^{p-2}+6^{p-2}-1$ is divisible by $p$ by Fermat Small Theorem ($a^{p-2}\equiv 1/a$, while $1/2+1/3+1/6-1=0$). For $n=2$ we have $a_n=48$ and so it is divisible by 6.

Another very easy problem, two others are not much harder. I predict cutoff for a Gold around 35 points.

Mine solution is similarly to yours: if n is prime then by li'l Fermat we have: a n-1=2^(n-1)+ 3^(n-1) + 6^(n-1) - 1 leaves remainder 2 (mod n), for n>=3. (*) But for n=3 we have a2=48 which is divisible by 3 and becase of (*) we have contradiction. Therefore, there is no prime solution so there is no solution at all. I hope that i didn't made some stupid mistake.

Fedor Petrov wrote: It is not locked? Participants may not read our messages? Thank you for rapidity, Valentin! For this problem, I think answer is only $1$ and it is clear, since for any prime $p>3$ we have $2^{p-2}+3^{p-2}+6^{p-2}-1$ is divisible by $p$ by Fermat Small Theorem ($a^{p-2}\equiv 1/a$, while $1/2+1/3+1/6-1=0$). For $n=2$ we have $a_n=48$ and so it is divisible by 6. Same solutions as yours, except that I wrote it a bit differently because I'm not used to write stuff like $2^{p-2}\equiv\frac{1}{2}$ because it doesn't seem ok (I mean $2^3\equiv\frac{1}{2}$ would mean that 5 divides 15/2 from the definition of congruence).. So I wrote it like this $2^{p-2}=\frac{px+1}{2}, 3^{p-2}=\frac{py+1}{3}, 5^{p-2}=\frac{pz+1}{5}$ and then we get $a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1=\frac{3px+3+2py+2+pz+1}{6}-1=\frac{3px+2py+pz}{6}$ which is obviously divisible by p (for p>=5), and beacuse $a_1=10$ and $a_2=48$ we see that there's no solution in prime numbers..

No, it looks OK, though may be not usual for one seing it first time. In a ring of residues modulo $n$ those residues coprime to $n$ are invertible, i.e. if $a$ and $n$ are coprime, then there exists $a'$ such that $a\cdot a'\equiv 1$ modulo $n$. So, if we write $b/a$, we mean $b\cdot a'$.

idioteque wrote: Same solutions as yours, except that I wrote it a bit differently because I'm not used to write stuff like $2^{p-2}\equiv\frac{1}{2}$ because it doesn't seem ok (I mean $2^3\equiv\frac{1}{2}$ would mean that 5 divides 15/2 from the definition of congruence).. So if you work with rationals you should interpret "p/q=0 mod m" as "m divides p and is coprime to q".

Strange ! I think this is the easiest Number theory problem in IMO . The most difficult part : Predict with each prime $ p >3 $ : $ 2^{p-2}+3^{p-2}+6^{p-2} -1 $ is divisible by $ p $ (It's proved easily by Fermat Little Theorem) I think the 2nd day is easier than the 1st day

For any prime p>3 x=2^{p-2}, y=3^{p-2}, z=6^{p-2} Since 2x\equiv 3y\equiv 6z (\equiv 1) (\mod p) and \gcd(p,2)=\gcd(p,3)=1 We have x\equiv 3z, y\equiv 2z (\mod p) So x+y+z\equiv 6z\equiv 1 (\mod p) or p|a_{p-2}

My solution is very similar to yours:

I've got another (more or less) nice solution: The main idea is to make a recursion of that explicit term for $a_n$. It is given that $a_n=2^n+3^n+6^n+(-1)*1^n$, so the 4 solutions for the characteristic equality $q^4+A*q^3+B*q^2+C*q+D$ are $1, 2, 3, 6$ We still have to calculate the coefficients A, B, C, D which is done easily by Vieta: $D=1*2*3*6=36$ $C=-(1*2*3+2*3*6+1*3*6+1*2*6)=-72$ $B=1*2+1*3+1*6+2*3+2*6+3*6=47$ $A=-(1+2+3+6)=-12$ We therefore get $a_{n+4}=12a_{n+3}-47a_{n+2}+72a_{n+1}-36a_n$ By calculating the first 4 values for $a_n$ by the explicit term, we get $a_1=10$ $a_2=48$ $a_3=250$ $a_4=1392$ And now we can calculate (and this time without any concerns about calculating with rational numbers) $a_0=2$ and $a_{-1}=0$. As $a_n$ is unique in both directions modulo any prime, 0 will appear again. Therefore 1 is the only solution. Yimin

thatw as also the solution I gave in the contest...

Yimin Ge wrote: As $a_n$ is unique in both directions modulo any prime, 0 will appear again. Therefore 1 is the only solution. Yimin modulo any prime not dividing $36$!!!!!!!!!

The number $a_n=2^n+3^n+6^n -1$, $n\geq 1$ is a complex number so we are done

complex I mean that it is not an odd number

silouan wrote: The number $a_n=2^n+3^n+6^n -1$, $n\geq 1$ is a complex number so we are done ?????? why are we done??? What about odd integers?

we find $ 2^n+3^n+6^n-1=(3^n+1)(2^n+1)-2 $

I think it's just a easy problem in IMO. 6*a_p-2 is divisible by p with p>=3, it can be solved by Ferma Little Theorem, and a₂ divisible by 2,3. That's all. Oh, I see it's smiliar to yours

[Mine?

silouan wrote: we find $ 2^n+3^n+6^n-1=(3^n+1)(2^n+1)-2 $ When I was at IMO I also wrote that in this way and solved it by FlT. But it seems that you solved it only using the parity of the numbers. (even or odd). So can you detail your solution?

Consider primes $p \neq 2,3$. Note that $p=2$ and $p=3$ can be manually checked not to work for $n=2$. Else, the sequence is clearly periodic every $p-1$ terms, so we can consider negative numbers $n$. For $n=-1$, \[a_{-1} = \frac{1}{2}+\frac{1}{3}+\frac{1}{6}-1 = 0.\] Thus, for every prime $p$, there exists some $n$ such that $p \mid a_n$, so the only value possible is $\boxed{1}$.

It's crazy you can do this. $\newline$ We can prove that for some arbitrary $p$, there is a $n$ so that \[2^n + 3^n + 6^n - 1 \equiv 0\pmod{p}\]Let $n \equiv -1 \pmod{p-1}$. Then \[\frac{1}{2} + \frac{1}{3} + \frac{1}{6} - 1 \equiv 0\pmod{p}\]so $a_n$ is divisible by every prime at some point. Hence, our answer is $1$.

It is enough to find a prime number p p/an= pow(2,n) + pow(3,n) + pow(6,n) -1 a2= 2² + 3² + 6² - 1= 4+9+36-1=48 p=2, p=3 For p>=5 From Fermat's little theorem; Pow(2, p-1)== pow(3, p-1) == pow( 6, p-1) (mod p) 6* ( pow( 2, p-2) + pow( 3, p-2) + pow( 6, p-2) -1)==0 (mod p) So, p / 6* a(p-2) , also (p;6)=1 so, p / a(p-2)

If we find all primes such that $p \not \mid a_{i}$ for all $i$, we can construct every set with them. However, note that \[a_{p-2} \equiv \frac{1}{2}+\frac{1}{3}+\frac{1}{6}-1 \equiv 0\pmod{p}\]for all $p\neq 2, 3$. It's easy to see that $2$ and $3$ divides $a_1$ and $a_2$ respectively. Thus, the only such $n$ is $1$.

We claim that the sequence $\{a_n\}$ contains all primes. Consider a prime $p$. If $p=2$ then we have $3^n \equiv 1 \mod(2)$ which is true for any $n$. If $p=3$ then we have $2^n \equiv 1 \mod(3)$ which is true for all even $n$. So from now on we consider only $p$ such that $\gcd(6,p)=1$. Thus if $p$ divides some $a_n$ then we have $2^n+3^n+6^n-1 \equiv 0 \mod(p) \implies 2^n+3^n+6^n \equiv 1 \mod(p)$. Note that since $\gcd(6,p)=1$, we can multiply both sides by the inverse of $6^n \mod(p)$ to get $( \frac{1}{3} )^n+\big (\frac{1}{2} \big)^n \equiv 0 \mod(p) \implies ( -\frac{2}{3} )^n \equiv 1 \mod(p)$. Since $p$ is coprime to $2,3$, there exists an order for $-\frac{2}{3} \mod (p)$, thus we may take $n$ to be that order and we are done. This means only 1 is relatively prime to all terms of the sequence.

We claim that $1$ is the only positive integer relatively prime to all terms of $a_n$. Since this is obviously true, consider some positive integer $m \geq 1$. If $m$ is a power of $2$ or $3$, then we can take $n=2$ to get $a_2 = 2^2 + 3^2 + 6^2 - 1 = 48$ which is not relatively prime to $m$. Otherwise, consider a prime $p \neq 2,3$ that divides $m$. Taking $n=p-2$, we have \begin{align*} a_{p-2} &= 2^{p-2} + 3^{p-2} + 6^{p-2} - 1 \\ &= 2^{p-1} \cdot 2^{-1} + 3^{p-1} \cdot 3^{-1} + 6^{p-2} \cdot 6^{-1} - 1 \\ &\equiv 2^{-1} + 3^{-1} + 6^{-1} - 1 \pmod p \, \text{(By Fermat's Little Theorem)} \\ &\equiv 0 \pmod p \end{align*}so $\gcd(a_{p-2}, m) \geq p > 1$, implying that there always exists at least $1$ term of $a_n$ that shares a prime factor with every positive integer greater than $1$, as claimed. $\blacksquare$

Me struggling to understand how you are doing this

The answer is just $1$, which clearly works. Manually verify that for $n = 2$, $3$ is a prime divisor, and for $n = 1$, $2$ is a prime divisor. For all other primes $p$, take $n = p - 2$, then we have $2^{p - 2} + 3^{p - 2} + 6^{p - 2} \equiv \frac 12 + \frac 13 + \frac 16 - 1 = 0 \mod p$, so no primes are relatively prime to all elements of the sequence, done.

The answer is $\boxed{1}$, which clearly works. We will prove that there are no primes $p$ relatively prime to all $a_n$. There are two cases. $p = 2$, $3$: Here, we can take $a_2=48$. $p\ne 2$, $3$: Here, we will prove that $$a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1\equiv 0\pmod p.$$Note that $$2^{p-2}+3^{p-2}+6^{p-2}-1\pmod p$$$$\equiv\frac{2^{p-1}}{2}+\frac{3^{p-1}}{3}+\frac{6^{p-1}}{6}-1\pmod p$$$$\equiv \frac{1}{2}+\frac{1}{3}+\frac{1}{6}-1 \pmod p$$$$\equiv 0\pmod p,$$which completes the problem.

The answer is just $n=1$ which obviously works. It clearly suffices to show that $\forall p, \exists n \in \{ 1, 2, \dots \}, p|a_n$. The conclusion for $p=2$ is true for all $n$ and for $p=3$ it is true for all even $n$. For other $p$, consider $n=p-2$. So, $a_{p-2} = \frac12 + \frac13 + \frac16 - 1 \equiv 0 \pmod{p}$, so we are done.

Consider a prime $p>3$ and $n=p-2$. We have that $2^n +3^n +6^n -1 \equiv 0.5+0.333...+0.16... -1 \equiv 0$ modulo $p$ so all primes greater than $3$ divide atleast one term of this sequence so we are done. $2$ obviously divides all terms and $3$ divides all those with $n$ even.

Note that we can take $a_2=48$ so multiples of $2, 3$ don't work. Now, suppose that $p>3$ is a prime. Then by FLT $$a_{p-2} \equiv \frac12+\frac13+\frac16-1 \equiv 0 \pmod p.$$Therefore $1$ is the only solution.

Really easy problem. It's really nice to look at the solutions from 2005 to the present and see how the solutions have evolved with time. Clearly $a_2=48$ so multiples of $2$ and $3$ do not satisfy the desired condition. Further, for all primes $p>3$, by Fermat's Little Theorem we have, \[2^{p-2}+3^{p-2}+6^{p-2}-1\equiv \frac{1}{2}+\frac{1}{3}+\frac{1}{6}-1 \equiv 0 \pmod{p}\]Thus, $p\mid a_{p-2}$ for all primes $p>3$. Thus, any integer with a prime factor greater than $3$ also violates the condition, implying that the only solution is $n=1$. Remark. It might be interesting to look at the following similar question. Harder Problem wrote: Determine all positive integers which have some multiple in the infinite sequence\[ a_n=2^n+3^n+6^n -1,\ n\geq 1. \] A similar argument as above shows that all $n$ for which $\gcd(n,6)=1$ satisfy this condition, but the cases where $\gcd(n,6)>1$ are messier. In particular any positive integer of the form $15k,20k,21k,28k$ does not satisfy this condition.