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What is the least number of digits (including repetitions) needed to e

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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
[Reveal] Spoiler: OA
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New post 02 Dec 2009, 08:02
10^n is a decimal number with a 1 followed by n zeros.
So 10^100 will include 100 0's + 1 digit for 1 = 101
So the answer is C.
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New post 02 Dec 2009, 12:59
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This can easily be solved by looking at patterns of 10:

\(10^1=2 digits\)
\(10^2=3 digits\)
\(10^3=4 digits\)
\(10^4=5 digits\)
...so...
\(10^n=n+1 digits\)

thus:
\(10^{100}=101 digits\)
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New post 02 Dec 2009, 13:32
10^2 is 1 followed by 2 zeroes in decimal notation

10^3 is 1 followed by 3 zeroes in decimal notation

10^4 is 1 followed by 4 zeroes in decimal notation

Hence 10^100 is 1 followed by 100 zeroes in decimal notation

Number of digits is 1 + 100 zeroes and hence answer C.
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New post 03 Dec 2009, 04:26
Thanks guys. OA is indeed C.
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Re: What is the least number of digits (including repetitions) needed to e [#permalink]

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New post 26 Jan 2015, 21:10
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wolver123 wrote:
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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New post 18 Apr 2016, 23:14
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New post 28 Sep 2016, 10:07
10^(-1)×10^101= 10^100
Can also be written in decimal notation as 0.1×10^101 = 10^100
The answer is C
What is the least number of digits (including repetitions) needed to e   [#permalink] 28 Sep 2016, 10:07
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