Need help to determine the base of the numbers!

[vijayiyer]

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Hello all,

Can anyone help me out with this problem?

I need to determine the Base/Radix of the numbers for the following operations to be correct.

Question1:- 24+17 = 40
Question2: 54/4 = 13


Thanks a lot in advance

 

[vijayiyer]

Hello all,

Can anyone help me out with this problem?

I need to determine the Base/Radix of the numbers for the following operations to be correct.

Question1:- 24+17 = 40
Question2: 54/4 = 13


Thanks a lot in advance

 

[_Eduardo_]

Question1:- 24+17 = 40

I think what the "minus" is no included.

4 + 7 = x0 --> The base could be 4d+7d = 11d

If verify (2d*11d + 4d) + (1d*11d + 7d) = 4d*11d+0d (OK)

Then Base = 11d

Question2: 54/4 = 13

54= 13*4 ==> 3d*4d = x4d ==> 3d*4d - 4d = 8d could be the Base.

If verify (5d*8d + 4d) = (1d*8d + 3d)*3d (OK)

Then Base = 8d

 

[vijayiyer]

Thanks very much.. But i dont think such computations are employed in digital design.

well, i'm afraid.. thanks a lot for the help but i couldnt find it in any of the digital books.. If you know any please advise

Thanks once again for the help.

 

[_Eduardo_]

Thanks very much.. But i dont think such computations are employed in digital design.

I am not sure if I have understood it well.

This is a quick check algebra-less. If what you need is a procedure, the steps are something like that:

24+17 = 40

The general equation is: a1*base + a0 + b1*base + b0 = c1*base + c0
--> (c1-a1-b1)*base = a0+b0-c0
--> base = (a0+b0-c0)/(c1-a1-b1)


54/4 = 13 --> 54 = 4*13

The general equation is: a1*base + a0 = b0*(c1*base + c0)
--> (a1-b0*c1)*base = b0*c0-a0
--> base = (b0*c0-a0)/(a1-b0*c1)

In both cases, if the quotient not integer or negative then the equation have no solution.

With major quantity of digits the solutions are the roots of a polynomial of ndigit-1 degree .
Again, only positive integer solutions have meaning.

 

[kspalla]

Question 1 base is 11
Question 2 base is 8