Write a product of two numbers in non decimal base If $A=(b-1) (b-1) (b-1)$ and $B=(b-1) (b-1)$ are written in base $b$, what is $A\times B$ in base $b$? 
I've tried developing $A \times  B$ in decimal base but couldn't get this product back in base $b$.
Beside this, is there a general procedure to compute product in non decimal base.
Thanks in advance for your help,
 A: I assume you mean "what's the product of two numbers $n$ and $m$ whose digits are all equal to the largest digit you can write in that base?" (e.g. $666_7$ and $66_7$ - in base $7$)? 
Hint: what are the digits of $(n+1)\times(m+1) = nm + ((n+1) +(m+1) -1)$? 
In general, multiplication in all bases is the same as in base $10$: multiply the digits $1$ by $1$ as you'd do in base 10, but obviously with a different multiplication table, with the usual rules for carryover.
A: Your two numbers as the largest possible three digit and largest possible two digit number written in base $b$. I.e. they are $1000_b-1_b$ and $100_b-1_b$. Algebraically these are equal to $b^3-1$ and $b^2-1$ respectively. As such their product will be given by:
$$(b^3-1)(b^2-1)=b^5-b^3-b^2+1$$
This however has negatives in it so its hard to read it as a base $b$ number. Lets manipulate until it is all positive values multiplied by powers of $b$.
$$=(b-1+1)b^4-b^3-b^2+1$$
$$=(b-1)b^4+b^4-b^3-b^2+1$$
$$=(b-1)b^4+(b-2+2)b^3-b^3-b^2+1$$
$$=(b-1)b^4+(b-2)b^3+2b^3-b^3-b^2+1$$
$$=(b-1)b^4+(b-2)b^3+b^3-b^2+1$$
$$=(b-1)b^3+(b-2)b^3+(b-1+1)b^2-b^2+1$$
$$=(b-1)b^3+(b-2)b^3+(b-1)b^2+b^2-b^2+1$$
$$=(b-1)b^3+(b-2)b^3+(b-1)b^2+1$$
This is now a sum of positive numbers times a power of $b$ so we can read off the coefficients to give up the digits in base $b$. So the digits in your product are $b-1$, $b-2$, $b-1$, $0$, $1$. E.g. in base 10 it would be 98901, in base 4 it would be 32301, etc.
A: Let throw away the usual meanings of $8,9,10$ and replace them with $10$ means $b$, $9$ means simply $b -1$ and $8$ means $b-2$.  We assume $b \ge 3$ and so if we were dealing with base $7$ or base $32$ then $9$ and $8$ means "six" and "five" in base $7$, and $9$ and $8$ means "thirty-one" and "thirty" in base $32$. 
Okay.  So  
$A = (b-1)(b-1)(b-1) = 999 =  b^3 -1 = 1000 - 1$
$B = (b-1)(b-1) = 99 = b^2 - 1 = 100 - 1$
So $AB = (b^3 -1)(b^2 - 1) = (1000 -1)(100 - 1)$
$100000 - 1000 - 100 + 1 = $
$99000 - 100 + 1 = $
$98900 + 1 = 98901 = (b-1)(b-2)(b-1)01$
This will be true for all $b \ge 3$
So $b=3: 21201$
$b=4: 32301$
$b=5: 43401$.....
$b=9: 87801$
$b=10: 98901$
$b=11: A9A01$.....
$b =16: FEF01$, etc.
