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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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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
[Reveal] Spoiler: OA

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

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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

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

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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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02 Dec 2009, 14:32
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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22 Oct 2010, 15:52
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, 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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23 Oct 2010, 18:49
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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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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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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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26 Jan 2015, 17:07
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Re: What is the least number of digits (including repetitions) needed to e [#permalink]

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26 Jan 2015, 22:10
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Expert's post
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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Re: What is the least number of digits (including repetitions) needed to e [#permalink]

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19 Apr 2016, 00:14
Hello from the GMAT Club BumpBot!

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

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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

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What is the least number of digits (including repetitions) needed to e   [#permalink] 28 Sep 2016, 11:07
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