The first step you need to take is to find out how many digits youare going to need. Hexadecimal is base 16 so we can startwith \(\displaystyle{16}^{{{0}}}={1}\) and \(\displaystyle{16}^{{{1}}}={16}\) and \(\displaystyle{16}^{{{2}}}={256}\) and \(\displaystyle{16}^{{{3}}}={4096}\) since \(\displaystyle{16}^{{{3}}}\) isgreater then the number we need we know that the largest digit is amultiple of 16 squared. First we take our number in decimaland divide it by 256 only using the whole number and discarding anydecimal. For our first digit we get 5 since \(\displaystyle{\frac{{{1457}}}{{{5}}}}={5.69140625}\).

Now we work with the remainder. to get theremainder take \(\displaystyle{1457}-{256}\times{5}={177}\). Note that 177 is less then 256 sothis is a valid number. We now want to look at the seconddigit which is held by multiples of \(\displaystyle{16}^{{1}}.{\frac{{{177}}}{{{16}}}}={11.0625}\) soour second digit is 11 or B. Again we need to find the remainder so \(\displaystyle{177}-{16}\times{11}={1}\) and since one is less then 16 again it is valid. Now for our digit in the \(\displaystyle{16}^{{{0}}}\) power since our remainder was 1 thatis our third digit. So far we have 5B1 but we need to look at forthe fractional part of the number which is the .11 in base 10 or \(\displaystyle{\frac{{{11}}}{{{100}}}}\).

In hex we our decimal point is \(\displaystyle\frac{{1}}{{16}}\) and \(\displaystyle{\frac{{{1}}}{{{16}^{{{2}}}}}}\) or \(\displaystyle{16}^{{-{1}}}\) then \(\displaystyle{16}^{{-{2}}}\) for two decimalplaces. So we need to know how many 16ths we havewith \(\displaystyle{\frac{{{11}}}{{{100}}}}\). To do this we simple multiply .11 by 16 giving 1.76 so the first decimal point over is 1 and the remainderis \(\displaystyle{\frac{{{11}}}{{{100}}}}-{\frac{{{1}}}{{{16}}}}={.048}\) now with the now wee need to know what is howmany times \(\displaystyle{16}^{{-{2}}}\) goes into .048 multiplying .048 by \(\displaystyle{\frac{{{1}}}{{{\left({16}^{{-{2}}}\right)}}}}={12}\).288 so the next digit is 12 or C if we wanted more numbers wewould continue by finding how many times \(\displaystyle{16}^{{-{{3}}}}\) goes into theremainder of .\(\displaystyle{048}-{12}\times{16}^{{-{2}}}\) but since the problem only wanted twoplaces we are done getting a final answer of 5B1.1C.

Part B is allot easier then A since binaryis a multiple of hex that is base 2 goes into base 16 or \(\displaystyle{2}^{{{4}}}={16}\) soall we need to do is convert each number directly intobinary. For the first digit "5" we replace it with the binary5= 0101 then B=1011 and 1= 0001 and 1=0001 and C=1100 thentake the numbers 5 and B and 1 and put them together givingfollowed by a decimal point then the numbers 1 and C giving 010110110001.00011100

for base 2 or binary. Now for Octal which is base 8 we can achieve this by convertingfrom binary directly since \(2^3=8\) all we need to do is groupthe numbers in groups of 3 first working to the rightfrom the decimal point gives 000 and 111 which becomes 0 and 7 soto the right of the decimal point we will have .070 the last zerocomes in since we don't have anything further right then 111 nowlooking from the left of the decimal point and working right toleft we get 001=1 then 110=6 then 110=6 then 010=2 however wedid that we worked right to so we flip the conversion giving 2661for what is to the left of the decimal point now combining the twoparts gives 2661.070

in octal. A neat way ofchecking your answers is to use the windows calculator and turn onall the features in the veiw menue by using the scientificfunction then typing in the number when decimal is selected thenclicking on the base you wish to go to and then it will displayyour number in the new base. I hope this has helped you.