How do I find the radix r of equalities?

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To find the radix r of equalities, one must express each term in the equation in terms of powers of r. For example, in the equation 14_r + 52_r + 3_r = 113_r, each number can be expanded to show its base r representation. The equation can be rewritten as (1*r^2 + 4*r) + (5*r + 2) + 3 = (1*r^2 + 1*r + 3). Solving this equation leads to the conclusion that r equals 6, while also noting that negative solutions are not valid in this context. Understanding the concept of radix is crucial for solving such problems effectively.
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how do i find the radix r of equalities.
for example how would i find the radix r of
14r + 52r +3r =113r

note:::: the r is a little subscribt on the bottom.
 
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Do you understand that the radix is the base being used to represent these numbers? In other words, that radix 10 means base 10 -- our common numbering system?

The number 125 in base 10 can be broken down to
1 \times 10^2 + 2 \times 10^1 + 5 \times 10^0

You can set up a system of equations to represent your equality in that form and solve them to discover the radix.

PS: next time you should post your homework questions in the homework forum.
 
ok then what would the r be in the above problem
 
In my example, the "r" is 10.

In your example, "r" is what you have to figure out. You don't want me to deny you all that fun, do you?

I'll get you started, though. You have the number "113" in some base r. Let me see you write that number in the same format as I used to write my 125, only where I used 10, you use r.
 
so 113=1X10^2 + 1X10^1 + 3X10^0 so in this case r also = 10.
but how doe 14r +52r + 3r = 113r?

im sorry for asking so many questions, but my prof literraly doent speak english and to top it off he stutters for about 5 sec on every word. and to top that off he's in his 80s.
 
You missed my point. In your 113_r, r is NOT 10.

But just as I split up 125_{10} into a sum of powers of 10, you can split up 113_r into a sum of powers of r. And you can similarly split up the other numbers in your equation. See what you can do with that.
 
(1*r+4)+(5*r+2)+3=1*r2+1*r+3

Solve for r and make sure r>5.
 
so would r be 6 or -1 in that case.
 
Do you think it could be -1?
 
  • #10
14r + 52r +3r =113r[\sub]

is exactly the same as
r+ 4+ 5r+ 2+ 3= 1r2+ r+ 3 in base 10. Can you solve that for r?
 
  • #11
so in this example r is equal to 6 right
 
  • #12
Yeah, but do you understand the concept that just because -1 is one of the solutions to the quadratic equation doesn't mean that it is a valid solution to the problem you were solving?
 
  • #13
yes. Thanks to anyone who spent time helping me out. I really appericiate it. :smile:
 

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