How many digits 2^100 has?

Answer = A = 31

\(2^{100} = 2^{10*10}\)

\(= (2^{10})^{10}\)

\(= 1024^{10}\)

\(= (1000 + 24)^{10}\)

\(= (10^3 + 24)^{10}\)

\(= 10^{30} + 10 * 10^{29} * 24 + ...........\)

Answer should be slight greater than 30

= 31

Answer = A

hmm....am sure there is a better way to do this, but -

2^10 = 1.024 * 10^3 => 2^100 = (1.024)^10 * 10^30

therefore 31 digits would be my best guess

cool way to do it, i didn't think about this one

I did it like this

at max 1 digit can be added after every third power, eg 2 4 8, 16 32 64, 128 etc
so max possible digit based on this observation can be (100 -1)/3 + 1 = 34
so the answer 31 comes naturally.

I solved it like sparky did - 2^100 = (2^3)^33 * 2

Number of digits is less than 34. So A is the only choice by elimination.

2^2=1 digit
2^3=1 digit
2^4=1 digit
2^5=2 digit
2^6=2 digit
2^7=3 digit
2^8=3 digit
2^9=3 digit
2^10=4 digits - so 2^20 will have 7 digits which is 3 digits difference -> 2^30 will have 10 digits and so on.. so 2^100 will have 31 digits

PareshGmat

Can you elaborate as to how (10^3+24)^10 equates to the last step.

Let me try to explain.

\((10^3+24)^{10}\) , by binomial expansion of \((a+b)^n\) = \(a^n+a^{n-1}*b......+a*b^{n-1}+b^n\)

Similarly,

\((10^3+24)^{10} = 10^{30}+....+10*24^9+24^{10}\)

\(10^{29}*24+.....+24^{10}\) will have digits of the same order as that for \(10^{30}\) to reach a total of 31 digits.

Options C-E are way too much for this question and hence are easily eliminated.
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The characteristic of the logarithm of a positive number is positive and it is one less than the number of digits in the number.

Take logarithm of the given number:-
=log(2^100)
=100*log(2)
=100*0.301
=30.1
hence total number of digits:-
=30+1
31

Bunuel
2^1 = 1 digit
2^2 = 1 digit
2^3 = 1 digit
2^4, 2^5, 2^6 has 2 digits
2^7, 2^8, 2^9 has 3 digits
there is a repetition of 3.
is there a way to solve this question through this approach?

my answer comes out to be 34 though....
100/3 +1
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The mistake you are doing is that you are assuming that the 'cyclicity' is actually 3. This is wrong

2^6 = 2 digits, 2^7 to 2^9= 3 digits BUT 2^10 to 2^14 = 4 digits (this breaks the cyclicity).

Thus, you can not use cyclicity for this question.

Easiest way is to use log function as mentioned above (but GMAT doesnt want you to know how log function works. If you do know it, becomes very straightforward). If not, use binomial theorem as mentioned in my post how-many-digits-2-100-has-17192.html#p1579232\

Alternately, you can solve it as :

2^100 = (2^10)^10 = 1024^10 = (1.024)^10* (1000)^10 = (a value just slightly greater than 1)*10^30, giving you a number with 31 digits.

Hope this helps.
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yeah i used the log function for this as well.

Very quick and easy

quickly Log2^100 is 100Log2 = 100* Log(2)


I would remember a few common logs in case a question like this pops up

Log(2) = 0.305
Log(3)=0.477
Log(4)=Log(4^2)=2Log(2)
Log(5)=0.698
Log(6)=Log(2*3)=Log(2)=Log(3)
Log(7)=0.845

Now, as we can seem any simple common base we can use this trick for, unless they give us something a bit more complicated like 13 or 17, then I am out of tricks. But this is good for numerous ways this question can be asked.

First of all, your AP is 1, 4, 7, 10, 13 ... 100.
So number of terms will be (100 - 1)/3 + 1 = 34. It will not be 100/3 + 1

Next what you need to understand here is that the 34 gives you the maximum number of digits that \(2^{100}\) can have.
Look at it this way. 128 -> 256 -> 512 -> 4 digit number
But this is the quickest that the number of digits can change. For some digits, you could actually have 4 numbers.
1024 -> 2048 -> 4096 ->8192 -> 5 digit number

Fortunately, the only option that is less than 34 (but more than 25) is 31. Hence it must be the answer.
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Looking through the solutions, logarithms appear to be the fastest and most accurate way to solve this problem assuming you've memorized some basic logarithms. I haven't been a member long enough or posted enough to include a link but if you search youtube for a video titled "Logarithms Example - 4 / Find The Number Of Digits using Logarithms - Maths Arithmetic" you can see a good example of using logarithms do determine the number of digits in an exponent that would take far too long to compute manually.

I don't think this question is something that the GMAT writers will test on the exam, unless someone here can confirm that such a question has appeared on the official GMAT. To the best of my knowledge, it does not follow the rules of how and what GMAT test writers like to test.

Here is an example of a question that is tested on the GMAT:

The number of digits in \((8^{16}) (25^{25})\) (when written in the decimal form) is .....

Cheers,
Dabral

First of all, your AP is 1, 4, 7, 10, 13 ... 100.
So number of terms will be (100 - 1)/3 + 1 = 34. It will not be 100/3 + 1

Karishma kindly explain this step.

How do you find the number of terms in an AP?

Number of terms (n) = (Last term - First term)/Common difference + 1

It is derived from

Last term = First term + (n - 1) * Common difference

So it will be
n = (100 - 1)/3 + 1 = 34

Check out this post for more on AP formulas:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html/2012/03 ... gressions/
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I solved it this way -

\(2^{100} = (1000 + 24)^{10}\).... Now, when you see the first term i.e. 1000 and its power 10.. the answer should lie close to 30.... 35 isn't possible because \(24^{10}\) will always be less than \(1000^{10}\).

Bunuel where can I find more questions like this?

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Hope it helps.
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