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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 e  [#permalink]

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New post 02 Dec 2009, 08:19
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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
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Re: What is the least number of digits (including repetitions) needed to e  [#permalink]

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New post 02 Dec 2009, 13:59
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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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Re: What is the least number of digits (including repetitions) needed to e  [#permalink]

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New post 02 Dec 2009, 09:02
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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Re: What is the least number of digits (including repetitions) needed to e  [#permalink]

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New post 02 Dec 2009, 14:32
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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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Re: 600-700 digits problem  [#permalink]

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New post 22 Oct 2010, 15:52
rtaha2412 wrote:
I had trouble interpreting and then solving this problem. Please advise

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


I have corrected the question. It should say \(10^{100}\)

This number is 1 followed by 100 zeroes, so you have 101 digits in all, hence the answer (c)
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Re: 600-700 digits problem  [#permalink]

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New post 23 Oct 2010, 09:03
Consider \(10^n\), n being 1, 2, 3 ........

\(10^2\) = 100 => 3 digits
\(10^3\) = 1000 => 4 digits .... so on

number of digits for \(10^n\) = n+1

\(10^100\)=> number of digits = 100+1 = 101

Correct option C
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Re: 600-700 digits problem  [#permalink]

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New post 23 Oct 2010, 18:49
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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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Re: 600-700 digits problem  [#permalink]

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New post 25 Oct 2010, 00:24
C is the ans

10^2 = 100
10^3 =1 000
10^4 = 1 0000
so 10 ^100 willm be having 1 hunderd zeros = total digits = 1+ 100 = 101
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Re: 600-700 digits problem  [#permalink]

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New post 25 Oct 2010, 19:45
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Hi utin,
Decimal notation is the writing of numbers in a base-10 numeral system. It derives its name from Deci = 10 I guess :)

examples = 55 = 10X5 + 5; 123 = (10^2)X1+(10^1)X2+(10^0)X3

This is the general numeric system used for expressing numbers.
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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, 22:10
1
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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What is the least number of digits (including repetitions) needed to e  [#permalink]

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New post 28 Sep 2016, 11:07
10^(-1)×10^101= 10^100
Can also be written in decimal notation as 0.1×10^101 = 10^100
The answer is C
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Re: What is the least number of digits (including repetitions) needed to e  [#permalink]

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