# Coverting denary numbers into octal, binary and hexadecimal form

I'm self studying (after not touching any match for a good 15 years) from Stroud & Booth's amazing "Engineering Mathematics". (This example is from page 55 of the 7th ediction, F.138 of the first chapter, "Arithmetic".)

The one thing I can't seem to get is the step by step conversion of a number (with decimals) from the denary form, into octal, binary, and hexadecimal form). I've got the denary to octal from down (step a), so I understand how to calculate:

$$348,654_{10} = 534,517_8$$

But the following steps are completely confusing to me.

(b) The authors take the octal form and write the binary equivalent of each digit in groups of three binary digits, thus:

$$101\:011\:100\: ,\: 101\:001\:111$$

This step I kinda understand, if I look at the binary number conversion tables, though would like to know how to do these manually. This bring us to: $$348,654_{10} = 534,517_8 = 101011100,101001111_{2}$$

But the next two steps are completely baffling to me.

(c) Starting from the decimal point, and working in each direction, the authors regroup the same binary digits in groups of four. This gives: $$0001\:0101\:1100\:,\:1010\:0111\:1000$$ completing the group at either end with extra zeros, as necessary.

Finally, they write out the hexadecimal equivalent of each group of four binary digits: $$1\:5\:(12)\:,\:(10)\:7\:8$$

Replacing (12) and (10) with their hexadecimal equivalents, the final result is: $$15C,A78_{16}.$$

Can anybody shine some light for me on this, especially steps (c) and (d). I'm completely stumped.