Hexadecimal Numbers
Hexadecimal numbers (also known as base-16) are a system of numbers which have 16 digits, instead of 10. These digits are \(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E\) and\(F\). \(A\) 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 learnt 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! For 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
\[721 \times 933 = 672693. \ _\square\]