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Write 82 in the base-two system.

[tex]$82 = \square$[/tex]

Answer :

To express [tex]$82$[/tex] in base two, we can use the division-remainder method. This involves dividing the number by [tex]$2$[/tex] repeatedly, keeping track of the remainders at each step, and then writing the remainders in reverse order.

Below are the detailed steps:

1. First Division:
Divide [tex]$82$[/tex] by [tex]$2$[/tex].
[tex]$$82 \div 2 = 41 \quad\text{with a remainder of } 0.$$[/tex]
Write down the remainder: [tex]$0$[/tex].

2. Second Division:
Now divide the quotient [tex]$41$[/tex] by [tex]$2$[/tex].
[tex]$$41 \div 2 = 20 \quad\text{with a remainder of } 1.$$[/tex]
Write down the remainder: [tex]$1$[/tex].

3. Third Division:
Divide [tex]$20$[/tex] by [tex]$2$[/tex].
[tex]$$20 \div 2 = 10 \quad\text{with a remainder of } 0.$$[/tex]
Write down the remainder: [tex]$0$[/tex].

4. Fourth Division:
Divide [tex]$10$[/tex] by [tex]$2$[/tex].
[tex]$$10 \div 2 = 5 \quad\text{with a remainder of } 0.$$[/tex]
Write down the remainder: [tex]$0$[/tex].

5. Fifth Division:
Divide [tex]$5$[/tex] by [tex]$2$[/tex].
[tex]$$5 \div 2 = 2 \quad\text{with a remainder of } 1.$$[/tex]
Write down the remainder: [tex]$1$[/tex].

6. Sixth Division:
Divide [tex]$2$[/tex] by [tex]$2$[/tex].
[tex]$$2 \div 2 = 1 \quad\text{with a remainder of } 0.$$[/tex]
Write down the remainder: [tex]$0$[/tex].

7. Seventh Division:
Finally, divide [tex]$1$[/tex] by [tex]$2$[/tex].
[tex]$$1 \div 2 = 0 \quad\text{with a remainder of } 1.$$[/tex]
Write down the remainder: [tex]$1$[/tex].

After reaching a quotient of zero, we collect all the remainders. The remainders in the order they were obtained (from the first division to the last) are:
[tex]$$0,\ 1,\ 0,\ 0,\ 1,\ 0,\ 1.$$[/tex]

Since the binary representation is formed by writing the remainders in reverse order, we reverse the list to get:
[tex]$$1,\ 0,\ 1,\ 0,\ 0,\ 1,\ 0.$$[/tex]

Thus, the binary representation of [tex]$82$[/tex] is:
[tex]$$1010010.$$[/tex]

To confirm, we can also convert it back to decimal:
[tex]$$1 \times 2^6 + 0 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 0 \times 2^0$$[/tex]
[tex]$$= 64 + 0 + 16 + 0 + 0 + 2 + 0 = 82.$$[/tex]

Therefore, the final answer is:

[tex]$$82 = 1010010.$$[/tex]

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