Which of the following binary (base-2) numbers is LARGEST?

A. 11000000
B. 01111111
C. 00000001
D. 10111111

1 answer

To determine which of the given binary numbers is the largest, we can convert each binary number to its decimal (base-10) equivalent.

A. 11000000

  • \(1 \times 2^7 + 1 \times 2^6 + 0 \times 2^5 + 0 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 0 \times 2^0\)
  • = \(128 + 64 + 0 + 0 + 0 + 0 + 0 + 0 = 192\)

B. 01111111

  • \(0 \times 2^7 + 1 \times 2^6 + 1 \times 2^5 + 1 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0\)
  • = \(0 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 127\)

C. 00000001

  • \(0 \times 2^7 + 0 \times 2^6 + 0 \times 2^5 + 0 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 1 \times 2^0\)
  • = \(0 + 0 + 0 + 0 + 0 + 0 + 0 + 1 = 1\)

D. 10111111

  • \(1 \times 2^7 + 0 \times 2^6 + 1 \times 2^5 + 1 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0\)
  • = \(128 + 0 + 32 + 16 + 8 + 4 + 2 + 1 = 191\)

Now we can summarize the decimal values:

  • A: 192
  • B: 127
  • C: 1
  • D: 191

The largest binary number among the options is A. 11000000, which is equal to 192 in decimal.