Hexadecimal Numbers
Hexadecimal numbers (also known as base-16) are a system of numbers which have 16 digits, instead of 10. These digits are
\[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F,\]
where \(A\) is 10, \(B\) is 11, etc.
In hexadecimal, place value is determined by powers of 16, instead of 10. For example,
\[2A3_{16} = 2 \times 16^2 + 10 \times 16^1 + 3 \times 16^0 = 675_{10}.\]
Determine the value of \(8C91_{16}\) in base 10.
\(8C91_{16}\) is equal to
\[8 \times 16^3 + 12 \times 16^2 + 9 \times 16^1 + 1 \times 16^0 = 32768 + 3072 + 144 + 1 = 35985_{10}.\]
Therefore, the answer is \(35985.\) \(_ \square \)
Now that you have learned about hexadecimal numbers, give this problem a try:
Find the sum of \(3A5_{16}\) and \(2D1_{16}\) in base 16.
We will first convert these numbers to decimal (base 10) and work out the sum, and at the end we will convert back:
\[\begin{align} 3A5_{16} &= 3 \times 16^2 +10 \times 16^1 + 5 \times 16^0 = 768 + 160 + 5 = 933_{10}\\ 2D1_{16} &= 2 \times 256 + 13 \times 16^1 + 1 \times 16^0 = 512 + 208 + 1 = 721_{10}. \end{align}\]
Now we need to find the sum of \(933\) and \(721\), which is \(1654\).
Finally we convert back. Since \(1654 = 6 \times 16^2 + 7 \times 16^1 + 6 \times 16^0 = 676_{16},\) the answer is \(676_{16}.\) \(_ \square\)
However, with a lot of practice, the sum can also be done without changing the base back to 10. For instance,
Find the sum of \(3A5_{16}\) and \(2D1_{16}\) in base 16.
We have
\[\begin{align} 3A5+2D1 &=3A0+2D0+6\\ &=300+200+176\\ &=676. \end{align}\]
Note all numbers in this segment are in base 16. \(_\square\)
Multiplication will be complicated as instead of a \(10 \times 10\) multiplication chart, we have a \(16 \times 16\) one. Much harder!
Look at the following example:
Find the product of \(3A5_{16}\) and \(2D1_{16}\) in base 16.
We have
\[3A5 \times 2D1= 74A00+2F610+3A5=A43B5.\]
In base 10, we have just calculated
\[721 \times 933 = 672693. \ _\square\]