Question

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

Answer 1

`1001011_2`
`+`
`100011_2`
`=`
`6E_(hex)`

Step-by-step 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)`