Question

What is the output of the sum of 1001011 and 100011 displayed in hexadecimal?

Answer 1

Answer:

$1001011_2$    $+$   $100011_2$   $=$    $6E_{hex}$

Explanation:

Required

$1001011_2 + 100011_2 = []_{16}$

First, carry out the addition in binary

$1001011_2$    $+$   $100011_2$   $=$    $1101110_2$

The step is as follows (start adding from right to left):

$1 + 1 = 10$ --- Write 0 carry 1

$1 + 1 + 1(carry) = 11$ ---- Write 1 carry 1

$0 + 0 + 1(carry) = 1$ ---- Write 1

$1 + 0 = 1$ --- Write 1

$0 + 0 = 0$ ---- Write 0

$0 + 0 = 0$ ---- Write 0

$1 + 1 = 10$ --- Write 0 carry 1

No other number to add ; So, write 1 (the last carry)

So, we have:

$1001011_2$    $+$   $100011_2$   $=$    $1101110_2$

Next, convert $1101110_2$ to base 10 using product rule

$1101110_2 = 1 * 2^6 +1 * 2^5 + 0 * 2^4 + 1 * 2^3 + 1 * 2^2 + 1 * 2^1 + 0 * 2^0$

$1101110_2 = 64 +32 + 0 + 8 + 4 + 2 + 0$

$1101110_2 = 110_{10}$

Lastly, convert $110_{10}$ to hexadecimal using division and remainder rule

$110/16 \to 6\ R\ 14$

$6/16 \to 0\ R\ 6$

Write the remainder from bottom to top;

$110_{10} = 6(14)_{hex}$

In hexadecimal

$14 \to E$

So, we have:

$110_{10} = 6E_{hex}$

Hence:

$1001011_2$    $+$   $100011_2$   $=$    $6E_{hex}$