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.

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$8C91_{16}$ 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.

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We will first convert these numbers to decimal (base 10) and work out the sum, and at the end we will convert back:

$\begin{aligned} 3A5_{16} &= 3 \times 16^2 +10 \times 16^1 + 5 \times 16^0 = 768 + 160 + 5 = 933_{10}\\ 2D1_{16} &= 2 \times 256 + 13 \times 16^1 + 1 \times 16^0 = 512 + 208 + 1 = 721_{10}. \end{aligned}$

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.

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We have

$\begin{aligned} 3A5+2D1 &=3A0+2D0+6\\ &=300+200+176\\ &=676. \end{aligned}$

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.

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We have

$3A5 \times 2D1= 74A00+2F610+3A5=A43B5.$

In base 10, we have just calculated

$721 \times 933 = 672693. \ _\square$