# How to find the radix (base) of a number given its representation in another radix (base)?

What's the method to find the base of any given number?

E.g. find $r$ such that $(121)_r=(144)_8$, where $r$ and $8$ are the bases.

So how do I find the value of $r$?

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By definition $$144_8=1\cdot8^2+4\cdot8^1+4\cdot8^0=64+32+4=100\;,$$ and $$121_r=1\cdot r^2+2\cdot r^1+1\cdot r^0=r^2+2r+1\;.$$ To find $r$, just solve the quadratic equation $$r^2+2r+1=100$$ by whatever method you find most convenient; the slickest method is probably to notice that $r^2+2r+1=(r+1)^2$, so $(r+1)^2=100=10^2$.

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So value of r here would be i guess, 9 ? (r+1)^2= 100 -> r+1 = 10 -> r = 10-1 -> r =9 – Kishan Thobhani Oct 20 '12 at 12:22
@Kishan: That’s right. And you can now check: $$121_9=1\cdot9^2+2\cdot9^1+1\cdot9^0=81+18+1=100\;.$$ – Brian M. Scott Oct 20 '12 at 12:27
This doesn't uniquely describe the base though, we can also have $r+1 = -10$, which implies $r = -11$ which also works, and there's nothing saying a base has to be positive – CameronJWhitehead Feb 2 '15 at 14:04

First let's find what 144(8) equals, with "8" as the base Listing the digits in order, count them off from the right, starting with zero:

digits: 1 4 4 numbering: 2 1 0

Then the bottom row of numbers become the assigned exponential numbers, and so we have: 144= 1⋅8ˆ2 + 4⋅8ˆ1 + 4⋅8ˆ0 = 64 + 32 + 4 = 100

As for 121(r), with "r" as the base, we can again list off the digits in order, counting them off from the right starting with zero:

digits: 1 2 1 numbering: 2 1 0

Again, the bottom row of numbers become the assigned exponential numbers, and so we have: 121r= 1⋅rˆ2 + 2⋅rˆ1 + 1⋅rˆ0 = rˆ2 + 2r + 1

We can then find "r" by solving the quadratic equation that we have now created: rˆ2 + 2r + 1 = 100 equation (r + 1)ˆ2 = 100 = 10ˆ2 factor (r + 1) ± r + 1 = ± 10 use the square root property ± r = ± 10 - 1 isolate the "r" r = 9 solve for r

HOPE THIS HELPS!!! JUST CLARIFICATION OF ALL THE STEPS SO IT APPEARS LONGER THAN APPRECIATED :D

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for clarification – user82889 Jun 17 '13 at 21:57

remember the bold letters are the powers

(121)r=(144)8

Now r and 8 is the base so,

$1\cdot r^2+2 \cdot r^1+1\cdot r^0 = 1 \cdot 8^2+4\cdot 8^1+4 \cdot 8^0$

$r^2+2r+1=64+32+4$

$r^2+2r+1=100$

$(r+1) \textbf{2}=100$

$r+1=10$

$r=9$

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Essentially the same as Brian's answer, but hey, welcome to this site! You may want to learn how to format your equations, and avoid posting duplicate answers in the future. – FrenzY DT. Dec 23 '13 at 8:19