Assume that there is no number that is divisible by $3$ on the circle. It is clearly seen that if $a,b$ is any two neighboring numbers on the circle then $a,b$ have a different remainder when divided by $3$ because if $a-b \equiv 0 \pmod {3}$, a contradiction since $a-b$ is equal to $1$ or $2$ or $\frac ab=2$. Hence, if there are $n$ numbers that have a remainder of $1$ when divided by $3$ then there also have $n$ numbers on the circle that have a remainder of $2$ when divided by $3$. But this shows a contradiction since then, $n= \frac{99}{2}$. Thus, there exists a natural number on the circle such that is divisible by $3$.

your solution is not clear

you should say that is a number on circle is 3k+1 then it's neighbour is just 3k+2 ( if we assume that there is no number that is divisble by 3) beacuse if their devide is 2 beacause both are natural number then k should be odd because 3k+1 shoud be even then 3k+1=6m+4(k=2m+1) and 6m+4/2=3m+2 so it's neigbor is 3m+2 so the numbers on the circle is 3k+1,3k+2,3k+1,3k+2,.... and this is contradiction sorry for my bad english

HINT : If none of the numbers were a multiple of $3$ , then all these operations are simply methods of changing the residue modulo $3$.

let no number is divisible by $3$. $a_{i}-a_{i+1}=1$ or $2$ or $a_{i+1}$ that implies no two consecutive numbers have same residue $mod $3 . now select numbers with residue $\equiv 1(mod3)$ and $\equiv 2(mod3)$ one after one . there must be two consecutive numbers with same residue . so , it contradicts our assumption . $\therefore $ there is a number divisible by 3 .

Assume for the sake of contradiction that no number is divisible by 3. Reading indices modulo 3 if $a_i$ and $a_{i+1}$ are two adjacent nos. then $1 \equiv (a_i - a_{i+1} )^2 \text(mod 3)$ . And so by assumption , $a_i$ and $a_{I+1}$ are distinct modulo 3 , so they alternate between $0$ and $1$ . But this is impossible as we have an odd no. of numbers arranged around the circle . Hence our assumption was false and we are done.

Let us assume no number is divisible by $3$.Obseve each operation is a function to switch Modulo 3.(from 1 to 2 and viceversa).Two adjacent mos can't have same Modulo $3$.Hence every alternate number has same $(mod3)$.Thus the $1st$ and $99th$ no have same $(mod3)$. Thus they won't satisfy any of the conditions.Thus a Contradiction!

Let the only residues be $\{-1, 1\}$ mod $3$, then, since no two neighbouring numbers give same residue, these values must alternate. Since $99$ is odd, we get a contradiction!

anantmudgal09 wrote: Let the only residues be $\{-1, 1\}$ mod $3$, then, since no two neighbouring numbers give same residue, these values must alternate. Since $99$ is odd, we get a contradiction! Very nice solution.. but why does the question give the information that $\frac{a}b$ may be $2$.. it's never used.. even in this simple solution, you have not used it..

Beautiful problem Let's tag the $99$ numbers with $A_1, A_2, A_3, \ldots, A_{99}$ Claim: We are gonna have at least 50 even. Proof: Suppose, $A_k$ is odd. Then, $A_{k + 1} = A_k + 1$ or $2 \times A_k$. Therefore, $A_{k + 1}$ must be an even number. This proof is also true for the previous one of $A_k$. Using the previous claim, we get we have two consecutive even numbers. So, $A_p$ has $1 \pmod{3}$ and $A_{p + 1}$ has $2 \pmod{3}$ [We are considering the worst case.] Then, $A_{p + 2}$ must be a number having $1 \pmod{3}$. If we look at the sequence carefully we can see that for all $A_{p + 2k}$ will have $2 \pmod{3}$. And $99$ is odd, so the previous one of $A_p$ has $1 \pmod{3}$. But, is that possible? So, a contradiction. We will surely have a number divisible by $3$. Observation: For every odd $n$ that is true.

stranger_02 wrote: anantmudgal09 wrote: Let the only residues be $\{-1, 1\}$ mod $3$, then, since no two neighbouring numbers give same residue, these values must alternate. Since $99$ is odd, we get a contradiction! Very nice solution.. but why does the question give the information that $\frac{a}b$ may be $2$.. it's never used.. even in this simple solution, you have not used it.. without a/b = 2 it would have been too easy, but the key thingy is that multiplying by 2 switches the residue class mod 3 so alternate still exists.