Finding Unknown Bits in Binary Number Addition

The problem asks us to find the values of the binary digits \(x\) and \(y\) in the number \((10011x0y)_2\), such that the sum of three binary numbers \((11011)_2\), \((10110110)_2\), and \((10011x0y)_2\) equals the binary number \((101101101)_2\).

Understanding Binary Addition

Binary addition is performed column by column from right to left, similar to decimal addition. When adding multiple binary numbers in a column, the sum of the bits in that column, plus any carry from the previous column, determines the result bit for the current column and the carry to the next column.

For a given column \(i\), let \(b_{1,i}\), \(b_{2,i}\), and \(b_{3,i}\) be the bits from the three numbers at that position, and let \(c_{in}\) be the carry-in from the previous column. The total sum for the column is \(T = b_{1,i} + b_{2,i} + b_{3,i} + c_{in}\).

The bit for the result in column \(i\) is \(T \pmod 2\).

The carry-out to the next column (\(i+1\)) is \(\lfloor T/2 \rfloor\).

Aligning the Binary Numbers for Addition

To add the binary numbers, we align them by their rightmost digit (the \(2^0\) position). We can pad the shorter numbers with leading zeros to match the length of the sum, which is 9 bits in this case.

• \((11011)_2\) becomes \((000011011)_2\)
• \((10110110)_2\) becomes \((010110110)_2\)
• \((10011x0y)_2\) becomes \((010011x0y)_2\)

The sum is given as \((101101101)_2\).

Let's set up the column addition:

Step-by-Step Column Addition to Find x and y

We will perform binary addition column by column from right to left, solving for \(y\) and \(x\) as we encounter them.

Column \(2^0\) (Rightmost)

Bits from the numbers are 1, 0, and \(y\). Carry-in from the previous column is 0 (as it's the first column).

Sum of bits + carry-in = \(1 + 0 + y + 0 = 1 + y\).

The sum bit for this column is 1. Using the rule: \(1+y = 1 + 2 \times \text{Carry-out}\).

For the sum bit to be 1, \(1+y\) must be an odd number. This happens if \(y=0\).

• If \(y=0\): Sum is \(1+0 = 1\). \(1 = 1 + 2 \times \text{Carry-out}\) implies \(0 = 2 \times \text{Carry-out}\), so Carry-out \(C_1 = 0\). The result bit is \(1 \pmod 2 = 1\), which matches the target sum bit.
• If \(y=1\): Sum is \(1+1 = 2\). \(2 = 1 + 2 \times \text{Carry-out}\) implies \(1 = 2 \times \text{Carry-out}\), which is impossible for integer carry. The result bit is \(2 \pmod 2 = 0\), which does not match the target sum bit (1).

Therefore, \(y\) must be 0. Carry-out to the next column (\(2^1\)) is \(C_1 = 0\).

Column \(2^1\)

Bits are 1, 1, 0. Carry-in is \(C_1 = 0\).

Sum of bits + carry-in = \(1 + 1 + 0 + 0 = 2\).

The sum bit for this column is 0. \(2 = 0 + 2 \times \text{Carry-out}\) implies \(2 = 2 \times \text{Carry-out}\), so Carry-out \(C_2 = 1\). The result bit is \(2 \pmod 2 = 0\), which matches the target sum bit.

Carry-out to the next column (\(2^2\)) is \(C_2 = 1\).

Column \(2^2\)

Bits are 0, 1, and \(x\). Carry-in is \(C_2 = 1\).

Sum of bits + carry-in = \(0 + 1 + x + 1 = x + 2\).

The sum bit for this column is 1. Using the rule: \(x+2 = 1 + 2 \times \text{Carry-out}\).

For the sum bit to be 1, \(x+2\) must be an odd number. This happens if \(x=1\).

• If \(x=0\): Sum is \(0+2 = 2\). \(2 = 1 + 2 \times \text{Carry-out}\) implies \(1 = 2 \times \text{Carry-out}\), impossible. The result bit is \(2 \pmod 2 = 0\), does not match target (1).
• If \(x=1\): Sum is \(1+2 = 3\). \(3 = 1 + 2 \times \text{Carry-out}\) implies \(2 = 2 \times \text{Carry-out}\), so Carry-out \(C_3 = 1\). The result bit is \(3 \pmod 2 = 1\), which matches the target sum bit.

Therefore, \(x\) must be 1. Carry-out to the next column (\(2^3\)) is \(C_3 = 1\).

We have found \(x=1\) and \(y=0\). Let's verify the remaining columns using these values.

The third number is \((10011100)_2\).

Column \(2^3\)

Bits are 1, 0, 1. Carry-in is \(C_3 = 1\).

Sum of bits + carry-in = \(1 + 0 + 1 + 1 = 3\).

Result bit = \(3 \pmod 2 = 1\). Carry-out \(C_4 = \lfloor 3/2 \rfloor = 1\). Target sum bit is 1. Matches.

Carry-out to the next column (\(2^4\)) is \(C_4 = 1\).

Column \(2^4\)

Bits are 1, 1, 1. Carry-in is \(C_4 = 1\).

Sum of bits + carry-in = \(1 + 1 + 1 + 1 = 4\).

Result bit = \(4 \pmod 2 = 0\). Carry-out \(C_5 = \lfloor 4/2 \rfloor = 2\). Target sum bit is 0. Matches.

Carry-out to the next column (\(2^5\)) is \(C_5 = 2\).

Column \(2^5\)

Bits are 0, 1, 0. Carry-in is \(C_5 = 2\).

Sum of bits + carry-in = \(0 + 1 + 0 + 2 = 3\).

Result bit = \(3 \pmod 2 = 1\). Carry-out \(C_6 = \lfloor 3/2 \rfloor = 1\). Target sum bit is 1. Matches.

Carry-out to the next column (\(2^6\)) is \(C_6 = 1\).

Column \(2^6\)

Bits are 0, 0, 0. Carry-in is \(C_6 = 1\).

Sum of bits + carry-in = \(0 + 0 + 0 + 1 = 1\).

Result bit = \(1 \pmod 2 = 1\). Carry-out \(C_7 = \lfloor 1/2 \rfloor = 0\). Target sum bit is 1. Matches.

Carry-out to the next column (\(2^7\)) is \(C_7 = 0\).

Column \(2^7\)

Bits are 0, 1, 1. Carry-in is \(C_7 = 0\).

Sum of bits + carry-in = \(0 + 1 + 1 + 0 = 2\).

Result bit = \(2 \pmod 2 = 0\). Carry-out \(C_8 = \lfloor 2/2 \rfloor = 1\). Target sum bit is 0. Matches.

Carry-out to the next column (\(2^8\)) is \(C_8 = 1\).

Column \(2^8\)

Bits are 0, 0, 0. Carry-in is \(C_8 = 1\).

Sum of bits + carry-in = \(0 + 0 + 0 + 1 = 1\).

Result bit = \(1 \pmod 2 = 1\). Carry-out \(C_9 = \lfloor 1/2 \rfloor = 0\). Target sum bit is 1. Matches.

The final carry \(C_9\) is 0.

Conclusion

From the step-by-step binary addition process, we determined the values of \(x\) and \(y\).

The value of \(x\) is 1.

The value of \(y\) is 0.

This corresponds to the third number being \((10011100)_2\).

Revision Table: Binary Addition Steps for Finding x and y

Additional Information: Binary Numbers and Arithmetic

Binary numbers are a base-2 number system, using only the digits 0 and 1. Each position in a binary number represents a power of 2, increasing from right to left (\(2^0, 2^1, 2^2\), and so on).

Binary addition is a fundamental operation in digital electronics and computer science. Understanding how to add binary numbers, including handling carries, is essential.

When adding more than two binary numbers, the carry logic extends. If the sum of bits in a column plus the carry-in is \(T\), the result bit is \(T \pmod 2\) and the carry-out is \(\lfloor T/2 \rfloor\). For example, if the sum is 3 (decimal) in a column, the result bit is \(3 \pmod 2 = 1\) and the carry-out is \(\lfloor 3/2 \rfloor = 1\). If the sum is 4 (decimal), the result bit is \(4 \pmod 2 = 0\) and the carry-out is \(\lfloor 4/2 \rfloor = 2\) (which means a carry of 10₂ to the next column, equivalent to two standard carries of 1). This is why the carry values can be greater than 1 when adding more than two numbers simultaneously column by column.