recursion for the difference between 99 and 100

Question

Write a recursion for the difference between 99 and 100, as you move up in bases above 10 (base 11, base 12, base 13, etc.).
It should be fully simplified, and please use "n" (w/out quotes) as your variable. n should equal whichever base you're using (i. e. if you write a recursion and it happens to be n^2+2n+3, I should be able to plug in any base as n, and find out the difference between 99 and 100 for that base)

"Bases above 10? What's a base?" You may ask. Well, you would naturally count up in base 10, (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20....), base 11 however, would be (1,2,3,4,5,6,7,8,9,A,10,11,12,13,14,15,16,17,18,19,1A,20....), base twelve adding a B, and so on and so forth. I would look carefully at what happens between 99 and 100 though, it gets interesting.

 

Use a lowercase n
For something like "eight to the power of two" please write it as 8^2
It should be simplified down into a quadratic-esk equation (ax^2 + bx + c)

Answer 1

I believe the question is asking for a formula for 100_n - 99_n in terms on n.

 

You should remember that any number e.g. a four digit number with the following digits in base b

 

a = a₃ a₂ a₁ a_{0 (base b)} can be written as a linear combination of powers of b.

 

In this case a = a₀ + a₁b + a₂b² + a₃b³ .

 

Do this for 100_n  and  99_n and see where that leads you.

 

 

Any questions? Comment back.