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You use powers of 16...begin with the largest power that will fit...like in this case, 16^2 (256)
so, 273-256=17. U used 16^2 once, so ur number will be 1__ Now 16^1, 17-16=1; (number currently 11_), and lastly 16^0 will finish that last 1...soo, 111 >>
Yeah that's how I do it, work from left to right, you have to work out what the biggest power of 16 that will fit into 273 is first though.
Like he said in this case it's 16^2 so you have a 3 digit number.
So you subtract 16^2 from 273, which subtracts once, leaving 17.
So your leftmost digit is 1
Then you subtract 16^1 from 17, which subtracts once, leaving 1.
So your middle digit is 1 (because it subtracted once)
Then you subtract 16^0 (anything to the 0 is the number 1, but I wrote it 16^0 for form), which subtracts once, leaving 0.
So your right digit is 1.
You have 0 left so you stop.
The other way is to work from right to left, so you don't need to know how many digits you have before you start, but you have to divide.
273/16 = 17 remainder 1
1 is your rightmost digit.
17/16 = 1 remainder 1
1 is your second rightmost digit
1/16 = 0 remainder 1
1 is your third digit from the right
Your quotient was 0 so you stop.
I like the first way better myself, but this way works to, and is probably less work.
I'm not really sure why I like the left to right method better...
Guess cuz my calculator won't do integer division for me