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

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Intern
Joined: 08 Aug 2009
Posts: 17
What is the least number of digits (including repetitions) needed to e  [#permalink]

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02 Dec 2009, 07:19
1
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Difficulty:

15% (low)

Question Stats:

69% (01:02) correct 31% (01:11) wrong based on 203 sessions

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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
Intern
Joined: 31 Oct 2009
Posts: 27
Re: What is the least number of digits (including repetitions) needed to e  [#permalink]

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02 Dec 2009, 12:59
7
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:
$$10^{100}=101 digits$$
##### General Discussion
Manager
Joined: 19 Nov 2007
Posts: 179
Re: What is the least number of digits (including repetitions) needed to e  [#permalink]

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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
Intern
Joined: 11 Nov 2009
Posts: 9
Re: What is the least number of digits (including repetitions) needed to e  [#permalink]

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02 Dec 2009, 13:32
1
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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22 Oct 2010, 14:52
rtaha2412 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

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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23 Oct 2010, 08:03
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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23 Oct 2010, 17:49
1
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 $$100 + 1 = 101$$

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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Manager
Joined: 25 Aug 2010
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24 Oct 2010, 23:24
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
Manager
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25 Oct 2010, 18:45
1
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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Location: Pune, India
Re: What is the least number of digits (including repetitions) needed to e  [#permalink]

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26 Jan 2015, 21:10
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

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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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
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Re: What is the least number of digits (including repetitions) needed to e  [#permalink]

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10 Feb 2018, 14:49
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Re: What is the least number of digits (including repetitions) needed to e &nbs [#permalink] 10 Feb 2018, 14:49
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