Mirror Image

Let's express our initial binary number as $(a,b,c,d,e)$ , where each letter represents either 1 or 0. The value of this number is therefore $16a+8b+4c+2d+e$ . The number as it appears in the mirror is this number flipped, so we reverse the letters and get the value of this new number as $16e+8d+4c+2b+a$ . Since this number is half of the original number, we can write $16a+8b+4c+2d+e=2(16e+8d+4c+2b+a)$ . Multiplying out the bracket, subtracting one side from the other and tidying things up gives us $31e+14d+4c=4b+14a$ . Remembering that each letter represents either 1 or 0, we can see that $e$ must be 0, since (among other other reasons) if $e$ were 1, $31e$ would be odd, and $14d+4c$ and $4b+14a$ can only ever be even, so the sides cannot be equal. Removing this gives the rather symmetrical $14d+4c=4b+14a$ . We can ignore the letters and focus on the numbers by removing the number if its accompanying letter represents a 0, and keeping it if the letter is 1. The question becomes: how many ways can we safely remove numbers from $14+4=4+14$ , keeping the equation correct? Clearly, we could remove the fours, the fourteens, or no numbers at all- giving us 11110, 10010 and 01100. Thus, the answer is $\boxed{3}$ .

Let the binary string be ABCDE , on reflection it will be EDCBA . Now we are given that the mirror image is half of the original number, hence if we right shift original number by 1 i.e 0ABCD is what should be the mirror image. This gives us the condition 0ABCD = EDCBA i.e E= 0, A=E, B=C , On trying out all combinations, there are only 3 non-zero solutions, hence answer is 3.

Let the binary string be $\overline{abcde}$ , then its mirror image is $\overline{edcba}$ .

It is required that $\overline{abcde}$ is twice the value of $\overline{edcba}$ . This further implies $\overline{abcde}$ is divisible by 2 (or $10$ ) and hence $e=0$ .

Giving, $\overline{abcde}=\overline{abcd0}$ and $\overline{0dcba}=\overline{abcd}$ . (Dividing $\overline{abcde}$ by 2.)

Now it is clear to see that the conditions are $a=d,b=c$ . There are a total of 4 possibilities. However, one of them would correspond to $a=b=c=d=0$ , which is not needed. Hence, there are a total of $\boxed{3}$ such numbers.