Question

question 24 of 25
based on the information in this table, how would the hexadecimal number
77a6 be written in binary code?
hexadecimal | binary
4 | 0100
5 | 0101
6 | 0110
7 | 0111
8 | 1000
9 | 1001
a | 1010
b | 1011
c | 1100
a. 111011110100110
b. 1000101111000101
c. 1000101100010111
d. 100011010111001

Explanation:

Step1: Break hex into digits

Hex number is 7, 7, A, 6.

Step2: Find binary for each digit

- 7: 0111 (from table)
- 7: 0111
- A: 1010 (from table)
- 6: 0110 (from table)

Step3: Combine the binaries

Combine 0111, 0111, 1010, 0110 → 0111011110100110? Wait, no, wait: Wait, maybe I missed leading zeros? Wait, no, each hex digit is 4 bits. Wait 7 is 0111 (4 bits), 7 is 0111, A is 1010, 6 is 0110. So combining: 0111 0111 1010 0110. But wait, maybe the options have leading 1s? Wait no, maybe I made a mistake. Wait the hex number is 77A6. Let's recheck:

Wait 7 in binary is 0111 (4 bits), 7 is 0111, A is 1010, 6 is 0110. So concatenate: 0111011110100110. But option A is 111011110100110. Wait, maybe leading zero is omitted? Wait 0111 is 7, but if we write as 111 (3 bits) no, hex to binary is 4 bits per digit. Wait maybe the original hex is 77A6, which is four digits: 7 (4th digit), 7 (3rd), A (2nd), 6 (1st)? No, hex is written left to right, so 7 (first digit) is 0111, 7 (second) 0111, A (third) 1010, 6 (fourth) 0110. So combining: 0111 0111 1010 0110. But option A is 111011110100110. Wait, maybe I messed up the digit order? Wait no, maybe the hex number is 77A6, which is 7,7,A,6. Let's check the options:

Option A: 111011110100110. Let's split into 4-bit chunks (from right? No, left to right). Wait 1110 1111 0100 110? No, that's not 4 bits. Wait maybe the hex number is 77A6, which is 7 (0111), 7 (0111), A (1010), 6 (0110). So concatenated: 0111011110100110. But option A is 111011110100110. Oh! Wait, maybe the first digit 7 is written as 111 (3 bits) but no, hex to binary is 4 bits. Wait no, 7 in binary is 0111 (4 bits), but maybe the leading zero is dropped? Wait 0111 is 7, but if we write as 111, that's 3 bits, but hex to binary requires 4 bits per digit. Wait maybe the problem has a typo, or I misread. Wait the hex number is 77A6. Let's check the binary for each digit again:

7: 0111 (from table: 7 is 0111)

7: 0111

A: 1010 (from table: A is 1010)

6: 0110 (from table: 6 is 0110)

So combining: 0111 0111 1010 0110 → 0111011110100110. Now, option A is 111011110100110. Wait, if we remove the leading zero from the first 4 bits: 0111 → 111? No, that's 3 bits. Wait no, 0111 is 4 bits. Wait maybe the hex number is 77A6, which is 7 (0111), 7 (0111), A (1010), 6 (0110). So concatenated: 0111011110100110. Now, let's count the bits: 4+4+4+4=16 bits. Option A is 15 bits? Wait no, let's count: 1 1 1 0 1 1 1 1 0 1 0 0 1 1 0 → 15 bits? No, 111011110100110 is 15 digits? Wait no, 1110 (4) 1111 (4) 0100 (4) 110 (3)? No, that's not. Wait maybe I made a mistake in the digit order. Wait hex number 77A6: let's write each digit as 4 bits:

7: 0111

7: 0111

A: 1010

6: 0110

So combining: 0111011110100110 (16 bits). Now, option A is 111011110100110 (15 bits). Wait, maybe the first digit is 7 (0111) but written as 111 (3 bits) and then the rest? No, that's incorrect. Wait maybe the hex number is 77A6, which is 7 (0111), 7 (0111), A (1010), 6 (0110). So concatenated: 0111011110100110. Now, let's check the options:

Option A: 111011110100110 → Let's split into 4-bit chunks from the right: 110 (3), 0100 (4), 1111 (4), 1110 (4). No, that's not. Wait maybe the hex number is 77A6, and the binary is formed by each digit's binary:

7: 0111

7: 0111

A: 1010

6: 0110

So concatenated: 0111011110100110. Now, let's compare with option A: 111011110100110. Oh! Wait, 0111 is 7, but if we invert the first bit? No, that doesn't make sense. Wait maybe the table has 7 as 0111, which is correct. Wait maybe the hex number is 77A6, and the binary is 0111011110100110, which is the same as option A but with a leading zero? Wait option A is 111011110100110, which is 0111011110100110 without the leading zero? Wait no, 0111011110100110 is 16 bits, option A is 15 bits. Wait, maybe I made a mistake in the digit. Wait hex number 77A6: 7 (0111), 7 (0111), A (1010), 6 (0110). So concatenated: 0111011110100110. Now, let's write this as a binary number: 0111011110100110. Now, let's remove the leading zero: 111011110100110, which is option A. Yes! Because leading zeros in binary numbers are not written, so 0111011110100110 is the same as 111011110100110 (removing the first zero). So that's option A.

Answer:

A. 111011110100110