Conversion from base 10 to other bases is a little bit trickier than converting to base 10.  The method we use is similar to the method above in that we take very good notice of the column headings and their respective values but use them in a slightly different way.


To convert from base 10 to another base we look at the largest column heading that will divide into the base 10 number, and make a note of how many times it does so and the remainder that is left.  We then take the remainder and divide into it the next lowest power of the new base, once again we make a note of how many times this works and once again the remainder which is then used as the next dividend for the next lowest power.


Yes… I can hear you saying to yourself “what is he talking about?”,  Perhaps if we take a look at an example it will start to make some sense.


Consider the base 10 number 423, suppose that we wish to convert this to base 3, how do we approach this problem?


Let us take a look at the column headings for base 3:



Step 1 - make a note of the highest base 3 column heading that does not exceed the number that were trying to convert.  In this case it is 243.


Step 2 - using 243 as our divisor and 423 as a dividend, carry out the long division:

243 goes into 423 only once, with the remainder of 180 shown in red in the usual “primary school” notation.


Remember that 243 was the largest base heading but we have to make the division using everyone below it.


Step 3 - this time using 81 as our divisor, and 180 as our dividend we ask ourselves “how many times does 81 go into 180?”  And once again we perform a small long division to find out the answer.



We see that 81 goes into 180 twice with a remainder of 18, once again we show this as above in red.


Step 4 - the next heading that we have to consider is 27, and we ask ourselves “how many times will 27 go into 18?”  Well, quite clearly the answer is 0 and we have to take account of this by putting a 0 in the answer and once again making a note of the remainder:



Step 5 - our next base 3 heading is 9, and we ask ourselves “how many times will 9 go into 18, well I hope I don’t insult your intelligence with the long division this time, but have to show it anyway and we can say that it is 2 with the remainder of 0:

Step 6 - we still have 2 more base headings to go through, 3 and 1, but as you can see a reminder this time is 0 and it’s never going to change from that, even so we still to make the divisions so ask ourselves two questions:


“How many times will 3 go into 0?” - Answer 0

“How many times will 1 go into 0?” - Answer 0


And we append both of these 0 to the end of our answer, which becomes 120200.


So, we can say that the base 10 number 423, in base 3, is 120200:


In all of this, the most important step is to realise that once you’ve identified your first divisor (remember, this is the largest column heading that will go into the number without exceeding it), you must then divide all of your subsequent remainders sequentially and where the result of that division is 0 but with still a remainder you append 0 to the answer and carry on with the remainder untouched until you have divided by the final column heading which would be the “units”.  This sounds more complicated than it actually is, it’s true to say that in most subjects, mathematics being no exception it is harder to explain something than it is actually do it!


Let’s now take a look at another example, this time we will go the other way and convert a base 10 number into base 17, in other words a denary number into a septendecimal number:

Consider the decimal/denary number - 3887, that is three thousand eight hundred and eighty-seven.  What would this be as a septendecimal number?



Step 1 -write out the column headings in base 17 until you reach one which exceeds the number in question, disregard this one and use the ones you’ve got so far (refer to the table printed previously):



We can see that only 289, 17 and 1 will be use this time because the next column heading is 4913 is way above the number that we are using, 3887.

I’m not going to bother drawing up the long divisions this time, but I will just put into words the answers:


Step 1 - “how many 289’s go into 3887?”  = 13 r 130


Step 2 - “how many 17’s go into 130?” = 7 r 11


Step 3 - “how many 1’s go into 11?” = 11 r 0


Now we need to construct the answer, bearing in mind that base 17 will not only use 0 to 9 as digits, but will also use  ‘A’ through ‘G ‘ where A = 10, B = 11, C = 12, D = 13, E = 14, F = 15 and G = 16 respectively.


Our first result was 13 = D


Our second result was 7 = 7


Our third result was 11 = B


Our number, in base 17 is therefore D7B


You might be asking yourself if there is a way to convert directly from base to base, ie: from base 5 to base 13 etc without going through the intermediate step of base 10. The answer is that there does not seem to be an easy way to do this, if any way exists at all.


Many computer programs are available that can carry out such a base conversion and would be able to do this in a few milliseconds, but it is likely that the developer has created an algorithm (ie: a formula, or documented set of steps in computer code) that converts from base X to base Y via base 10 but without revealing the intermediate stages. 

Computers are able to conduct many billions of instructions per second so the complexities of the intermediate calculations would be of little impact to the processor.