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Is there any easy way to find how many digits will be there

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Is there any easy way to find how many digits will be there [#permalink] New post 21 Jan 2004, 21:50
Is there any easy way to find how many digits will be there in product such as 12^12 x 15^18? Can we at least find approximate number of digits?
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 [#permalink] New post 22 Jan 2004, 06:43
Here's an approach. It assumes familiarity with logarithms.
Also, it's handy to memorise log2 = 0.3, log3=0.48, log5=0.7
All logs are on a base of 10.

If x= 12^12 x 15^18
then
logx =log(12^12 x 15^18)
= log(3^12 x 4^12 x 3^18 x 5^18)
= log(2^24 x 3^30 x 5^18)
= log(2^24) + log(3^30) + log(5^18)
= 24log2 + 30log3 + 18log5
= 24x0.3 + 30x0.48 + 18x0.7
= 34.2
therefore
x = 10^34.2

And that means that x has 35 digits
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 [#permalink] New post 22 Jan 2004, 08:01
pusht wrote:
Here's an approach. It assumes familiarity with logarithms.
Also, it's handy to memorise log2 = 0.3, log3=0.48, log5=0.7
All logs are on a base of 10.

If x= 12^12 x 15^18
then
logx =log(12^12 x 15^18)
= log(3^12 x 4^12 x 3^18 x 5^18)
= log(2^24 x 3^30 x 5^18)
= log(2^24) + log(3^30) + log(5^18)
= 24log2 + 30log3 + 18log5
= 24x0.3 + 30x0.48 + 18x0.7
= 34.2
therefore
x = 10^34.2

And that means that x has 35 digits


Thanks, I never thought of applying logarithms since it is not tested on GMAT. But in any case this is one way to solve the problem and it is nice. I wonder if there is any other way to atleast approximate the number of digits (without using the logarithms)
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 [#permalink] New post 23 Jan 2004, 03:23
Here's another approach...

12^12 x 15^18
= 4^12 x 3^12 x 3^18 x 5^18
= 2^24 x 3^30 x 5^18
= 2^6 x 3^30 x 10^18
= 64 x (10^1/2)^30 x 10^18 (note 10^1/2 = 3.16, which can be
approximated to 3 without any loss of
accuracy, as far as estimating the number
of digits in the product is concerned)
= 64 x 10^15 x 10^18
= 64 x 10^33

which has 35 digits!!!!
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 [#permalink] New post 23 Jan 2004, 07:44
pusht wrote:
Here's another approach...

12^12 x 15^18
= 4^12 x 3^12 x 3^18 x 5^18
= 2^24 x 3^30 x 5^18
= 2^6 x 3^30 x 10^18
= 64 x (10^1/2)^30 x 10^18 (note 10^1/2 = 3.16, which can be
approximated to 3 without any loss of
accuracy, as far as estimating the number
of digits in the product is concerned)
= 64 x 10^15 x 10^18
= 64 x 10^33

which has 35 digits!!!!


Fantastic!! The crucial point is to consider 3 = sqrt(10). Thanks
  [#permalink] 23 Jan 2004, 07:44
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