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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

$\sqrt{(121)_{r}}=11_{r}$

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

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

$1+r=1+r$

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

D)
why 11 can not be the answer as it is greater than 2
Because it is satisfying radix greater than 2 and 11 is subset of greater than 2 therefore 11 is incomplete answer
+1 vote
As we all know radix should be greater than number therefore any value > 2 is right and best answer

D)