How to Find the Base in an Equations with Unknown Numbers

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To find the base A in the equation 24)A + 17)A = 40)A, convert each number from base A to decimal form. For example, 24)A translates to 2A + 4, 17)A to 1A + 7, and 40)A to 4A. Substitute these expressions into the equation to form 2A + 4 + 1A + 7 = 4A. Simplifying this leads to a solvable equation for A. This method effectively avoids trial and error by using algebraic substitution.
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How would you solve a problem like this where you have to find what base A is for this equation to be true? So in what base are these numbers.24)A + 17)A = 40)A

Not asking the answer for this particular question but how you'd solve it (other than trial and error obviously)
 
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A two digit number ab)A in base A equals A\cdot a + b so:
24)A = 2A + 4
17)A = 1A + 7
40)A = 4A + 0
Then substitute in the equation and solve for A.
 
Simple:

24_A = 2*A^1+4*A^0 = 2A + 4

Similarly, 17_A = 1A + 7 and 40_A = 4A + 0

Substitute and solve for A
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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