If you are given f(x)= m243, use table 2 to solve the problem.
Let us break up 243 into its 2n components.
First of all we can see from table 2 that 243 falls between 128 and 256. Therefore, we must subtract 128 from 243.
243-128= 115
115 is the remainder but 115 itself can be further broken down into other 2n components. 115 falls between 64 and 128 therefore we can subtract 64 from 115.
115-64=51
51 is the remainder but 51 itself can be further broken down into other 2n components. 51 falls between 32 and 64 therefore we can subtract 32 from 51.
51-32=19
19 is the remainder but it can be further broken down into other 2n components. 19 falls between 16 and 32 therefore we can subtract 16 from 19.
19-16=3.
3 is the remainder and since it is a single digit remainder, we stop our operation here.
Therefore the final equation will look like this
f(x)=m243 = m128×m64×m32×m16×m3
That is how the table can be used quickly to break down any number into its 2n components