Convert the decimal number 173 to binary.

Convert the decimal number 173 to binary. 2. Convert the binary number 11010010 to decimal. 3. Convert the hexadecimal number A5 to binary. 4. Convert the binary number 1011100 to hexadecimal.

The Correct Answer and Explanation is:

Let’s break down each conversion step-by-step.

1. Convert the decimal number 173 to binary:

To convert decimal 173 to binary, divide the number by 2 and keep track of the remainders.

• 173 ÷ 2 = 86 remainder 1
• 86 ÷ 2 = 43 remainder 0
• 43 ÷ 2 = 21 remainder 1
• 21 ÷ 2 = 10 remainder 1
• 10 ÷ 2 = 5 remainder 0
• 5 ÷ 2 = 2 remainder 1
• 2 ÷ 2 = 1 remainder 0
• 1 ÷ 2 = 0 remainder 1

Now, write the remainders from bottom to top:

173 (decimal) = 10101101 (binary)

2. Convert the binary number 11010010 to decimal:

To convert the binary number 11010010 to decimal, multiply each digit by 2 raised to the power of its position, starting from 0 on the right.

• 1×27=1281 \times 2^7 = 1281×27=128
• 1×26=641 \times 2^6 = 641×26=64
• 0×25=00 \times 2^5 = 00×25=0
• 1×24=161 \times 2^4 = 161×24=16
• 0×23=00 \times 2^3 = 00×23=0
• 0×22=00 \times 2^2 = 00×22=0
• 1×21=21 \times 2^1 = 21×21=2
• 0×20=00 \times 2^0 = 00×20=0

Now, sum them up:

128 + 64 + 16 + 2 = 210

So, 11010010 (binary) = 210 (decimal).

3. Convert the hexadecimal number A5 to binary:

Each hexadecimal digit corresponds to a 4-bit binary number.

• A (hex) = 1010 (binary)
• 5 (hex) = 0101 (binary)

So, A5 (hexadecimal) = 10100101 (binary).

4. Convert the binary number 1011100 to hexadecimal:

Group the binary digits into sets of four starting from the right:

• 1011100 → 0010 1110 (add leading zeros to the left to make the group of 4 bits)

Now, convert each group to hexadecimal:

• 0010 (binary) = 2 (hex)
• 1110 (binary) = E (hex)

So, 1011100 (binary) = 2E (hexadecimal).

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Summary:

• 173 (decimal) = 10101101 (binary)
• 11010010 (binary) = 210 (decimal)
• A5 (hexadecimal) = 10100101 (binary)
• 1011100 (binary) = 2E (hexadecimal)

Each number system has its own rules, but the process involves understanding the place values for each number base (binary, decimal, hexadecimal).