The fastest way to answer this question is to list them out. The numbers have to be very small; the smallest 4-digit number in base 3 is 27; the largest 4-digit number in base 2 is 15.
n n (base 2) 2n 2n (base 3)
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1 1 2 2
2 10 4 11
3 11 6 20
4 100 8 22
5 101 10 101
6 110 12 110
7 111 14 112
8 1000 16 121
...
So the only two positive integers n for which n in base 2 has the same digits as 2n in base 3 are 5 and 6.
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However, there is an opportunity for some good mathematics in solving the problem formally.
We know that the leading digit is always 1, and that all the digits are either 0 or 1
(1) If the representation is a single digit, then it is the digit 1; in both base 2 and base 3 that represents the number 1; and that does not satisfy the requirement that the number represents n in base 2 and 2n in base 3.
(2) If the representation is two digits, then it is 1a, where a is either 0 or 1.
1a in base 2 is 2+a; 1a in base 3 is 3+a.
To satisfy the requirement that the number represents n in base 2 and 2n in base 3, we would need to have
So there are no 2-digit numbers that represent n in base 2 and 2n in base 3.
(3) If the representation is three digits, then it is 1ab, where both a and b are either 0 or 1.
1ab in base 2 is 4+2a+b; 1ab in base 3 is 9+3a+b.
To satisfy the requirement that the number represents n in base 2 and 2n in base 3, we would need to have
That is satisfied by ab=01 or by ab = 10. That gives us the two solutions we found earlier: 101 and 110.
101 base 2 = 5; 101 base 3 = 10 = 2(5)
110 base 2 = 6; 110 base 3 = 12 = 2(6)
(4) If the representation were four digits, then it would be 1abc, where a, b, and c are all either 0 or 1.
1abc in base 2 is 8+4a+2b+c; 1ab in base 3 is 27+9a+3b+c.
To satisfy the requirement that the number represents n in base 2 and 2n in base 3, we would need to have
Clearly that can not be satisfied by three integers a, b, and c which are all either 0 or 1.
So there are no numbers of four (or more) digits that represent the number n in base 2 and the number 2n in base 3.