Base of Number System: 121 = 324 Decimal

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The discussion revolves around determining the base of a number system where the representation "121" equals the decimal number "324." The problem can be expressed as a quadratic equation: 1*(b^2) + 2*(b) + 1 = 324. To solve for the base, one can use trial and error with values greater than 10. Understanding the conversion from base-b to decimal is crucial for finding the correct base. This approach clarifies the relationship between the two number representations.
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Homework Statement


What is the base of the system in which 121 represents the same number as the decimal number 324?


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The Attempt at a Solution


Can you please just explain to me what this means? I don't think we've learned it, and all the explanations I have found online are no help. I have no clue what the question is asking, but if someone could please reword perhaps what the question is asking, I'm sure I could figure it out. Thanks so much!
 
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We count using base 10. The decimal number 324 can be defined as follows:

3*(10^2) + 2*(10^1) + 4(10^0) = 324.

So,

324 = 1*(X^2) + 2*(X^1) + 1*(X^0)

This is now a quadratic equation which you can solve. The CORRECT resulting zero will be the base of the system in which the number 121 corresponds to 324 in base 10.
 
The digits in a number in the decimal (base-10) system represent increasingly higher powers of 10. The decimal number in this problem, 324, represents 3*102 + 2*101 + 4*100, or 300 + 20 + 4.

What this problem is asking for is the base b for which 1*b2 + 2*b + 1 is the same number as 324. One way to do this problem is trial and error - take a guess at what b might be, and see what 121b equal as a base-10 number.

Hint: Try values for b that are larger than 10.
 
Ok, thank you so much! This helped a lot.
 
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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