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Let $r$ denote number system radix. The only value(s) of $r$ that satisfy the equation $\sqrt{121_r}={11}_r$, is/are

1. decimal $10$
2. decimal $11$
3. decimal $10$ and $11$
4. any value > $2$
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$\sqrt{(121)_{r}}=11_{r}$

$\sqrt{(1\times r^{0})+(2\times r^{^{1}})+(1\times r^{2})}=(1\times r^{0})+(1\times r^{1})$

$\sqrt{(1+r)^{^{2}}}=1+r$

$1+r=1+r$

So any integer $r$ satisfies this but $r$ must be greater than $2$ as we have $2$ in $121$ and radix must be greater than any of the digits. (D) is the most appropriate answer

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+1
Please elaborate substitution at some extent. Means how option D is satisfying above relation
+3
Actually there is no need of any substitution.
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@Arjun sir plz make RHS also showing base r..plz edit the question..
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As we all know radix should be greater than number therefore any value > 2 is right and best answer

D)
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why 11 can not be the answer as it is greater than 2
+1
Because it is satisfying radix greater than 2 and 11 is subset of greater than 2 therefore 11 is incomplete answer
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is the second line the power of r is not correct by the way answer would not change

because in the unit place power of r should be 0

but due to symmetry answer will not change
+1 vote
As we all know radix should be greater than number therefore any value > 2 is right and best answer

D)
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