Answer :
Below is a step‐by‐step explanation to perform the binary multiplications.
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(a) Multiply
$$101_2 \times 11_2.$$
1. Convert each binary number to decimal:
- $$101_2 = 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 4 + 0 + 1 = 5.$$
- $$11_2 = 1 \times 2^1 + 1 \times 2^0 = 2 + 1 = 3.$$
2. Multiply the decimal values:
- $$5 \times 3 = 15.$$
3. Convert the product back to binary:
- $$15_{10} = 8 + 4 + 2 + 1 = 1 \times 2^3+ 1 \times 2^2+ 1 \times 2^1+ 1 \times 2^0,$$
- so $$15_{10} = 1111_2.$$
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(b) Multiply
$$111_2 \times 10_2.$$
1. Convert to decimal:
- $$111_2 = 4 + 2 + 1 = 7.$$
- $$10_2 = 2.$$
2. Multiply:
- $$7 \times 2 = 14.$$
3. Convert to binary:
- $$14_{10} = 8 + 4 + 2 = 1\cdot2^3+1\cdot2^2+1\cdot2^1+0\cdot2^0,$$
- so $$14_{10} = 1110_2.$$
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(c) Multiply
$$1011_2 \times 101_2.$$
1. Convert to decimal:
- $$1011_2 = 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 8 + 0 + 2 + 1 = 11.$$
- $$101_2 = 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 4 + 0 + 1 = 5.$$
2. Multiply:
- $$11 \times 5 = 55.$$
3. Convert to binary:
- The binary representation of $$55_{10}$$ is found by successively subtracting powers of 2:
- $$55 - 32 = 23$$ (so a 1 in the $$2^5$$ place),
- $$23 - 16 = 7$$ (1 in the $$2^4$$ place),
- $$7 - 0\ (since 8 > 7) \Rightarrow 0 \text{ in the } 2^3\text{ place},$$
- $$7 - 4 = 3$$ (1 in the $$2^2$$ place),
- $$3 - 2 = 1$$ (1 in the $$2^1$$ place),
- $$1 - 1 = 0$$ (1 in the $$2^0$$ place).
- Thus, $$55_{10} = 110111_2.$$
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(d) Multiply
$$1101_2 \times 10_2.$$
1. Convert to decimal:
- $$1101_2 = 1\cdot2^3 + 1\cdot2^2 + 0\cdot2^1 + 1\cdot2^0 = 8 + 4 + 0 + 1 = 13.$$
- $$10_2 = 2.$$
2. Multiply:
- $$13 \times 2 = 26.$$
3. Convert to binary:
- Express $$26_{10}$$ in binary:
- $$26 - 16 = 10$$ (1 in the $$2^4$$ place),
- $$10 - 8 = 2$$ (1 in the $$2^3$$ place),
- $$2 - 0\ (since 4 > 2) \Rightarrow 0 \text{ in the } 2^2\text{ place},$$
- $$2 - 2 = 0$$ (1 in the $$2^1$$ place),
- $$0 \text{ in the } 2^0\text{ place}.$$
- Therefore, $$26_{10} = 11010_2.$$
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(e) Multiply
$$1101_2 \times 101_2.$$
1. Convert to decimal:
- $$1101_2 = 13,$$
- $$101_2 = 5.$$
2. Multiply:
- $$13 \times 5 = 65.$$
3. Convert to binary:
- Decompose $$65_{10}$$:
- $$65 - 64 = 1$$ (1 in the $$2^6$$ place),
- $$1 - 0\ (next power 32 is too high) \Rightarrow 0 \text{ in lower places until }2^0,$$
- The remaining $$1$$ corresponds to $$2^0.$$
- Hence, $$65_{10} = 1000001_2.$$
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(f) Multiply
$$11010_2 \times 100_2.$$
1. Convert to decimal:
- $$11010_2 = 1\cdot2^4 + 1\cdot2^3 + 0\cdot2^2 + 1\cdot2^1 + 0\cdot2^0 = 16 + 8 + 0 + 2 + 0 = 26.$$
- $$100_2 = 4.$$
2. Multiply:
- $$26 \times 4 = 104.$$
3. Convert to binary:
- For $$104_{10}$$:
- $$104 - 64 = 40$$ (1 in the $$2^6$$ place),
- $$40 - 32 = 8$$ (1 in the $$2^5$$ place),
- $$8 - 8 = 0$$ (1 in the $$2^3$$ place),
- and zeros for the other positions.
- So, $$104_{10} = 1101000_2.$$
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(g) Multiply
$$10110_2 \times 110_2.$$
1. Convert to decimal:
- $$10110_2 = 1\cdot2^4 + 0\cdot2^3 + 1\cdot2^2 + 1\cdot2^1 + 0\cdot2^0 = 16 + 0 + 4 + 2 + 0 = 22.$$
- $$110_2 = 1\cdot2^2 + 1\cdot2^1 + 0\cdot2^0 = 4 + 2 + 0 = 6.$$
2. Multiply:
- $$22 \times 6 = 132.$$
3. Convert to binary:
- Find the binary form of $$132_{10}$$:
- $$132 - 128 = 4$$ (1 in the $$2^7$$ place),
- $$4 - 4 = 0$$ (1 in the $$2^2$$ place),
- Zeros in the other positions.
- Therefore, $$132_{10} = 10000100_2.$$
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(h) Multiply
$$10111_2 \times 101_2.$$
1. Convert to decimal:
- $$10111_2 = 1\cdot2^4 + 0\cdot2^3 + 1\cdot2^2 + 1\cdot2^1 + 1\cdot2^0 = 16 + 0 + 4 + 2 + 1 = 23.$$
- $$101_2 = 5.$$
2. Multiply:
- $$23 \times 5 = 115.$$
3. Convert to binary:
- Express $$115_{10}$$:
- $$115 - 64 = 51$$ (1 in the $$2^6$$ place),
- $$51 - 32 = 19$$ (1 in the $$2^5$$ place),
- $$19 - 16 = 3$$ (1 in the $$2^4$$ place),
- $$3 - 0\ (since 8 > 3) \Rightarrow 0 \text{ in the } 2^3\text{ place},$$
- $$3 - 2 = 1$$ (1 in the $$2^1$$ place),
- $$1 - 1 = 0$$ (1 in the $$2^0$$ place).
- Thus, $$115_{10} = 1110011_2.$$
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Final Answers:
a) $$101_2 \times 11_2 = 1111_2.$$
b) $$111_2 \times 10_2 = 1110_2.$$
c) $$1011_2 \times 101_2 = 110111_2.$$
d) $$1101_2 \times 10_2 = 11010_2.$$
e) $$1101_2 \times 101_2 = 1000001_2.$$
f) $$11010_2 \times 100_2 = 1101000_2.$$
g) $$10110_2 \times 110_2 = 10000100_2.$$
h) $$10111_2 \times 101_2 = 1110011_2.$$
This completes the step-by-step solution of the binary multiplications.
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(a) Multiply
$$101_2 \times 11_2.$$
1. Convert each binary number to decimal:
- $$101_2 = 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 4 + 0 + 1 = 5.$$
- $$11_2 = 1 \times 2^1 + 1 \times 2^0 = 2 + 1 = 3.$$
2. Multiply the decimal values:
- $$5 \times 3 = 15.$$
3. Convert the product back to binary:
- $$15_{10} = 8 + 4 + 2 + 1 = 1 \times 2^3+ 1 \times 2^2+ 1 \times 2^1+ 1 \times 2^0,$$
- so $$15_{10} = 1111_2.$$
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(b) Multiply
$$111_2 \times 10_2.$$
1. Convert to decimal:
- $$111_2 = 4 + 2 + 1 = 7.$$
- $$10_2 = 2.$$
2. Multiply:
- $$7 \times 2 = 14.$$
3. Convert to binary:
- $$14_{10} = 8 + 4 + 2 = 1\cdot2^3+1\cdot2^2+1\cdot2^1+0\cdot2^0,$$
- so $$14_{10} = 1110_2.$$
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(c) Multiply
$$1011_2 \times 101_2.$$
1. Convert to decimal:
- $$1011_2 = 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 8 + 0 + 2 + 1 = 11.$$
- $$101_2 = 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 4 + 0 + 1 = 5.$$
2. Multiply:
- $$11 \times 5 = 55.$$
3. Convert to binary:
- The binary representation of $$55_{10}$$ is found by successively subtracting powers of 2:
- $$55 - 32 = 23$$ (so a 1 in the $$2^5$$ place),
- $$23 - 16 = 7$$ (1 in the $$2^4$$ place),
- $$7 - 0\ (since 8 > 7) \Rightarrow 0 \text{ in the } 2^3\text{ place},$$
- $$7 - 4 = 3$$ (1 in the $$2^2$$ place),
- $$3 - 2 = 1$$ (1 in the $$2^1$$ place),
- $$1 - 1 = 0$$ (1 in the $$2^0$$ place).
- Thus, $$55_{10} = 110111_2.$$
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(d) Multiply
$$1101_2 \times 10_2.$$
1. Convert to decimal:
- $$1101_2 = 1\cdot2^3 + 1\cdot2^2 + 0\cdot2^1 + 1\cdot2^0 = 8 + 4 + 0 + 1 = 13.$$
- $$10_2 = 2.$$
2. Multiply:
- $$13 \times 2 = 26.$$
3. Convert to binary:
- Express $$26_{10}$$ in binary:
- $$26 - 16 = 10$$ (1 in the $$2^4$$ place),
- $$10 - 8 = 2$$ (1 in the $$2^3$$ place),
- $$2 - 0\ (since 4 > 2) \Rightarrow 0 \text{ in the } 2^2\text{ place},$$
- $$2 - 2 = 0$$ (1 in the $$2^1$$ place),
- $$0 \text{ in the } 2^0\text{ place}.$$
- Therefore, $$26_{10} = 11010_2.$$
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(e) Multiply
$$1101_2 \times 101_2.$$
1. Convert to decimal:
- $$1101_2 = 13,$$
- $$101_2 = 5.$$
2. Multiply:
- $$13 \times 5 = 65.$$
3. Convert to binary:
- Decompose $$65_{10}$$:
- $$65 - 64 = 1$$ (1 in the $$2^6$$ place),
- $$1 - 0\ (next power 32 is too high) \Rightarrow 0 \text{ in lower places until }2^0,$$
- The remaining $$1$$ corresponds to $$2^0.$$
- Hence, $$65_{10} = 1000001_2.$$
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(f) Multiply
$$11010_2 \times 100_2.$$
1. Convert to decimal:
- $$11010_2 = 1\cdot2^4 + 1\cdot2^3 + 0\cdot2^2 + 1\cdot2^1 + 0\cdot2^0 = 16 + 8 + 0 + 2 + 0 = 26.$$
- $$100_2 = 4.$$
2. Multiply:
- $$26 \times 4 = 104.$$
3. Convert to binary:
- For $$104_{10}$$:
- $$104 - 64 = 40$$ (1 in the $$2^6$$ place),
- $$40 - 32 = 8$$ (1 in the $$2^5$$ place),
- $$8 - 8 = 0$$ (1 in the $$2^3$$ place),
- and zeros for the other positions.
- So, $$104_{10} = 1101000_2.$$
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(g) Multiply
$$10110_2 \times 110_2.$$
1. Convert to decimal:
- $$10110_2 = 1\cdot2^4 + 0\cdot2^3 + 1\cdot2^2 + 1\cdot2^1 + 0\cdot2^0 = 16 + 0 + 4 + 2 + 0 = 22.$$
- $$110_2 = 1\cdot2^2 + 1\cdot2^1 + 0\cdot2^0 = 4 + 2 + 0 = 6.$$
2. Multiply:
- $$22 \times 6 = 132.$$
3. Convert to binary:
- Find the binary form of $$132_{10}$$:
- $$132 - 128 = 4$$ (1 in the $$2^7$$ place),
- $$4 - 4 = 0$$ (1 in the $$2^2$$ place),
- Zeros in the other positions.
- Therefore, $$132_{10} = 10000100_2.$$
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(h) Multiply
$$10111_2 \times 101_2.$$
1. Convert to decimal:
- $$10111_2 = 1\cdot2^4 + 0\cdot2^3 + 1\cdot2^2 + 1\cdot2^1 + 1\cdot2^0 = 16 + 0 + 4 + 2 + 1 = 23.$$
- $$101_2 = 5.$$
2. Multiply:
- $$23 \times 5 = 115.$$
3. Convert to binary:
- Express $$115_{10}$$:
- $$115 - 64 = 51$$ (1 in the $$2^6$$ place),
- $$51 - 32 = 19$$ (1 in the $$2^5$$ place),
- $$19 - 16 = 3$$ (1 in the $$2^4$$ place),
- $$3 - 0\ (since 8 > 3) \Rightarrow 0 \text{ in the } 2^3\text{ place},$$
- $$3 - 2 = 1$$ (1 in the $$2^1$$ place),
- $$1 - 1 = 0$$ (1 in the $$2^0$$ place).
- Thus, $$115_{10} = 1110011_2.$$
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Final Answers:
a) $$101_2 \times 11_2 = 1111_2.$$
b) $$111_2 \times 10_2 = 1110_2.$$
c) $$1011_2 \times 101_2 = 110111_2.$$
d) $$1101_2 \times 10_2 = 11010_2.$$
e) $$1101_2 \times 101_2 = 1000001_2.$$
f) $$11010_2 \times 100_2 = 1101000_2.$$
g) $$10110_2 \times 110_2 = 10000100_2.$$
h) $$10111_2 \times 101_2 = 1110011_2.$$
This completes the step-by-step solution of the binary multiplications.
