The famous Akbar and Birbal

Suppose Akbar gives Birbal some marbles leaving Akbar with $k$ marble and Birbal with $2k$ marbles. This implies that $n = k + 2k = 3k$ for some integer $k$ , i.e., $n$ is divisible by $3$ .

Similarly, after Birbal gives Akbar some marbles leaving Birbal with $m$ marbles and Akbar with $3m$ marbles, we have that $n = m + 3m = 4m$ for some integer $m$ , i.e., $n$ is divisible by $4$ .

Now the least positive integer divisible by both $3$ and $4$ is $12$ , and so $12$ is the minimum potential value for $n$ . Now say that the two start with $6$ marbles each. Then if Akbar gives Birbal $2$ of his marbles then Birbal will end up with $8$ marbles, twice the number Akbar would have. Also, If Birbal gives Akbar $3$ of his $6$ marbles then Akbar would have $9$ , thrice that of Birbal's $3$ marbles.

Thus we have a scenario where $n = 12$ is a possible solution, and since we've shown that $12$ is the minimum potential solution, we can conclude that $\boxed{12}$ is the minimum possible solution.

Assume, there are n no. of marbles. Initially, Akbar having x marbles . So, birbal get n-x marbles.

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1. A.Now , Akbar gives $a_{1}$ marbles to birbal

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$2(x-a_{1})=n-x+a_{1}$

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2 Now, Birbal gives $a_{2}$ marbles to akbar

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$x+a_{2}=3(n-x-a_{2})$

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3 From 1 and 2 after eleminating x ,

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$n=\frac{12(a_{1}+a_{2})}{5}$

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For n to be smallest natrural no.

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$a_{1}+a_{2}=5$

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$n=\boxed{12}$