How to convert 010111 to decimal using only odd numbered positions while still applying the correct positional exponent? I am trying to understand what this question is asking in terms of conversion. I know how to convert binary to decimal and vice versa, but the method you need to use has me confused.

I need to convert 010111 to decimal using only odd numbered positions while still applying the correct positional exponent in the conversion. What I did was take every odd numbered position from base 2 digit from 010111 so "111" I skipped every other number and took the digit from that position (if that makes sense) I only had three digits so I added a 0 to make it 4 bit 0111 which equals 7 in decimal. I know this is wrong but why??

This is the question: - Imagine that a processor is faulty and reads only odd-numbered positions when converting binary numbers to decimals, but it applies the correct positional exponent in the conversion. In radix 10, the binary number 010111 converted by this processor would be:

The mock test answer says the result is 2 but I cannot figure out why
I'm assuming positions are counted from $1$ starting from the least significant bit.

In this case you're using the exponents as if the number was 0111.
The question is asking to use the same exponents you would use for the odd-position digits of the number 010111.

To break it down, from the least to the most significant bit, the contribution to the final decimal value of each digit is:


First digit: $1 \cdot 2^0 = 1$.
Second digit: would be $1 \cdot 2^1 = 2$ but it is 0 since digits in even position are ignored.
Third digit: $1 \cdot 2^2 = 4$.
Fourth digit: ignored.
Fifth digit: $1 \cdot 2^4 = 16$.
Sixth digit: ignored.


The final result is: $16+4+1 = 21$.



Edit: Since you added that the answer should be $2$ this tells us that positions are either numbered starting from $0$, from the least to the most significant bit, or starting from $1$ from the most to the least significant bit.

In either case you would select the bits $001$ whose contribution would be $1 \cdot 2 + 0 \cdot 2^2 + 0 \cdot 2^4 = 2$.