No one on this site will know how do this. It's far too complicated, as for me I think you need to figure out how to do this yourself.
As per the 8421 rule the given binary number can be solved as : 0 0 0 0 0 1 1 1 0 0 0 0 0 4 2 1 i.e : 4+2+1 = 7 So the answer is 07 in decimal for 00000111 in binary
That all depends upon the first term of the sequence: As long as the first number is less than, or equal to 777 and as long as (first_number MOD 7) ≡ (777 MOD 7) will 777 be in the sequence. 777 MOD 7 ≡ 0 ⇒ if, and only if, first_number ≤ 777 and first_number MOD 7 ≡ 0 (ie 7 divides the first number) will 777 be in the sequence.
Binary numbers should start with 1 as 0 means that there is nothing in the column. So 00000111 is really just 111, which is the binary equivalent of the number 7
215
7 in Binary is 111
It all depends. You need 3 bits to get 8 potential States. 000 001 010 011 100 101 110 111 From top to bottom this counts from 0 to 7 in binary. In this example, the LSB for the sequence above would be 0, 1, 0, 1, 0, 1, 0 and 1. You really have to look at the overall decoding scheme. If you had a device with 7 inputs then you could assign each input to a specific segment. Depends on what you have to amswer your question precisely.
Decimal has ten different digits - 0 1 2 3 4 5 6 7 8 9 Binary only has two different digits - 0 1
To convert binary to hexadecimal split the binary number into blocks of 4 bits from the right hand end; each block represents a hexadecimal digit: 111101110001 → 1111 0111 0001 = 0xF71
The sequence from 3 to 7 can be described as consecutive integers.
No. As the sequence increases by 5 per term, the last digit repeats as "2" "7" "2" "7" etc. Clearly, "0" is never the last digit of any of those terms, and therefore, 200 will not be in the sequence.
7 - 3n: n = 0: 7-3(0) = 7 n = 1: 7-3(1) = 4 n = 2: 7-3(2) = 1 n = 3: 7-3(3) = - 2 n = 4: 7-3(4) = - 5 n = 5: 7-3(5) = - 8