Question 1161247
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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.<br><pre>
   n   n (base 2)  2n  2n (base 3)
  ---------------------------------
   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
  ...</pre>
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.<br>
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However, there is an opportunity for some good mathematics in solving the problem formally.<br>
We know that the leading digit is always 1, and that all the digits are either 0 or 1<br>
(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.<br>
(2) If the representation is two digits, then it is 1a, where a is either 0 or 1.<br>
1a in base 2 is 2+a; 1a in base 3 is 3+a.<br>
To satisfy the requirement that the number represents n in base 2 and 2n in base 3, we would need to have
{{{2(2+a) = 3+a}}}
{{{4+2a = 3+a}}}
{{{a = -1}}}<br>
So there are no 2-digit numbers that represent n in base 2 and 2n in base 3.<br>
(3) If the representation is three digits, then it is 1ab, where both a and b are either 0 or 1.<br>
1ab in base 2 is 4+2a+b; 1ab in base 3 is 9+3a+b.<br>
To satisfy the requirement that the number represents n in base 2 and 2n in base 3, we would need to have
{{{2(4+2a+b) = 9+3a+b}}}
{{{8+4a+2b = 9+3a+b}}}
{{{a+b = 1}}}<br>
That is satisfied by ab=01 or by ab = 10.  That gives us the two solutions we found earlier: 101 and 110.<br>
101 base 2 = 5; 101 base 3 = 10 = 2(5)
110 base 2 = 6; 110 base 3 = 12 = 2(6)<br>
(4) If the representation were four digits, then it would be 1abc, where a, b, and c are all either 0 or 1.<br>
1abc in base 2 is 8+4a+2b+c; 1ab in base 3 is 27+9a+3b+c.<br>
To satisfy the requirement that the number represents n in base 2 and 2n in base 3, we would need to have
{{{2(8+4a+2b+c) = 27+9a+3b+c}}}
{{{16+8a+4b+2c = 27+9a+3b+c}}}
{{{b+c = 11+a}}}<br>
Clearly that can not be satisfied by three integers a, b, and c which are all either 0 or 1.<br>
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.<br>