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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.
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
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
2^4 =16 it has a cycle of three.... i.e after every 3 powers, a digit increase ... but alas! it doesnt provide the correct answer
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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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kinaare paaon phailane lage hian, nadi se roz mitti kat rahi hai....
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
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.
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.
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
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 .....
\(2^{100} = (1000 + 24)^{10}\).... 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 \(1000^{10}\).
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gmatclubot
How many digits 2^100 has? &nbs
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31 Jan 2018, 19:52
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45% (01:22) correct 
