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Answer: C Explanation: The number of digits (least) for \(10^n = (n+1)\) Where, \(n \geq 0\).

So, The number of digits (least) for \(10^{100}\) are \(100 + 1 = 101\) Thus the answer is C.

Or, you can see that there are one hundred \(0's\) followed by \(1\) (to the right of \(1\)) and the least number of digits would be number of \(0's\) plus \(1\) ==> \(100+1=101\)
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26 Jan 2015, 17:07

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What is the least number of digits (including repetitions) needed to express 10^100 in decimal notation?

a) 4 b) 100 c) 101 d) 1000 e) 1001

Explanations please. Thanks

I am guessing what put you off the question was "in decimal notation". It just means using the decimal system of counting (with 10 digits 0 - 9) as opposed to other systems such as binary (using only two digits 0 and 1) etc. Decimal notation is the usual system we use to write numbers. It has nothing to do with the decimal point.
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Re: What is the least number of digits (including repetitions) needed to e [#permalink]

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19 Apr 2016, 00:14

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