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The positive integer a is a 2-digit number (01, 02 are not 2-digit number) the positive integer b has ‘a’ digit and the positive integer ‘c’ has ‘b’ digits. The smallest possible value for c is
Smallest number having $$n$$ digits is $$(10)^{n-1}$$
Smallest possible value of $$a=10$$
Now, $$b$$ has 10 digits, thus smallest possible value of $$b=(10)^9$$
Now, $$c$$ has $$(10)^9$$ digits, thus smallest possible value of $$c=(10)^{10^9-1}$$
=> Ans - (D)
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