Converting to Base 16; Place Value Chart

Date: 03/22/98 at 16:13:09
From: Terri
Subject: Base 16 numbers

How do you convert numbers to base 16 numbers?

Please use the following numbers in an example: 411213 and 38015.

Thank you.

Date: 03/22/98 at 20:01:39
From: Doctor Sam
Subject: Re: Base 16 numbers

Hi Terri,

Here's one way to think about other bases. If you were given 3 
hundred dollar bills and 4 ten dollar bills and 2 one dollar bills 
you would have 342 dollars.  We can just "glue" the 3 and the 4 and 
the 2 together because our decimal number system is based on tens.

Now if I gave you 3 quarters and 4 nickels and 2 pennies you wouldn't 
have 342 cents because nickels are only worth five pennies and 
quarters are only worth five nickels. But in base 5 notation that is 
just what you would write down. In base 5 notation every place value 
is five times as large as the one before it.  So 3 quarters, 4 
nickels, 2 pennies would be written as 342 base 5.

The key idea is that our system of writing numbers uses the idea of 
place value: that digits written in different places mean different 
amounts. 
 
Here is a chart of some place values in different bases:

   base 10:  ...    10*10*10      10*10    10    1    

    base 5:  ...    5 * 5 * 5     5 * 5     5    1

    base 2:  ...    2 * 2 * 2     2 * 2     2    1

   base 16:  ...    16*16*16      16*16    16    1

Now to your questions.

To change 411,213 begin by writing down the place values in base 16.  
The first few are:

     16^5       16^4       16^3       16^2      16^1      16^0
    1048576    65536       4096        256       16         1
 
411,213 < 1048576 so we will not use that place. But 411213 > 65536 so 
divide 65536 into 411213. The quotient is 6 and the remainder is 17997 
so there will be a 6 in the 16^4 place.

Now work with 17997.  17997 divided by 4096 has quotient 4 and 
remainder 1613 so there will be a 4 in the 16^3 place.

Now work with 1613.  1613 divided by 256 has quotient 6 and remainder 
77 so there will be a 6 in the 16^2 place.

Now work with 77.  77 divided by 16 has quotient 4 and remainder 13 so 
there is a 4 in the 16^1 place.

Finally, there is 13 remainder.  This is represented by a D in the 
ones place.

Here's how to organize the work.

   411213 = 6(17997) + 17997
    17997 = 4(4096)  + 1613
     1613 = 6(256)   + 77
       77 = 4(16)    + 13
       13 = D

Read down the quotients column to get the hex number 6464D.

To write this 38015 in hexadecimal do the same thing. Here's the work:

    38015 = 9(4096) + 1151
     1151 = 4(256)  + 127
      127 = 7(16)   + 15
       15 = F(1)

 So 38015 = 947F.

Your examples don't illustrate the following possibility, which may 
seem confusing.

To change 57536 to hex start again in the 4096 place (since 65536 is 
too large).

    57536 = 14(4096) + 192
 
Now 192 is smaller than the next hexadecimal place 256, but don't 
ignore that place. It will get a zero. To finish the problem:
  
    57536 = 14(4096) + 192
      192 =  0(256)  + 192
      192 = 12(16)  + 0
        0 =  0 (1)

Now 14 = E and 12 = C so 57536 = E0C0.

I hope that helps.

Doctor Sam,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/