quote:

Originally posted by creon444 Once again I need your help please. Suppose you have a a number in the form x^y where x is positive and y is relatively large (more than a hundred). Is there any way you could figure out (approximately) how many digits that number has without having to expand the whole thing first? Is it even possible? Like, for example, if your number is 5^800, can you tell how many digits it has just by looking at it? Please I need to know! Thanks to all of you!

well, here is a very ROUGH estimate...
5 ~= 2^2.. Thus 5^800 = (2^2)^800 = 2^1600.. Now 2^10 = 1024 ~= 1000.
Thus 2^1600 = (2^10)^160 ~= 1000 ^ 160 = 10 ^ 480. Now this has 480 + 1 = 481 digits. Thus a very rough estimate is 481 digits.

Last edited by Durafei on Feb-16-2004 at 23:14

common logarithm can be extremely useful in such situation.

with 5^800 i suppose u could do something like this

log(5^800) = 800log5 = 800log(10/2) = 800(log10 - log2) = 800-800log2.

now

800log2 = 80log(2^10) = 80log1024 > 80log1000 = 80*3 = 240

ie 800log2 > 240

also

80log1024-240 = 80(log1024-3) = 80(log1024-log1000) = 80log(1024/1000) = 80log(1000/1000+24/1000) = 80log(1+24/1000) = 80*24/1000*log(1+24/1000)^(1000/24)

now since limn→0(1+n)^(1/n) = e < 3

80*24/1000*log(1+24/1000)^(1000/24) < 80*24/1000*log3 = 48/25*log3 < 2log3 = log9 < 1

ie 80log1024-240 < 1 <=> 80log1024 < 241

evaluate

240 < 80log1024 < 241
240 < 800log2 < 241
-241 < - 800log2 < -240
-241+800 < -800log2+800 < -240+800
559 < 800-800log2 < 560
559 < log(5^800) < 560
log(10^559) < log(5^800) < log(10^560)
10^559 < 5^800 < 10^560

your number has _exactly_ 560 digits.

in general, u would need to use some arithmetic to estimate the value of logn first. once u've figured ou t the range, u should be able to get an answer. though it will be approximate in most cases...

___________________

quote:

your number has _exactly_ 560 digits.

LOL - my estimate was just "slightly" wrong

For some reason I totally forgot that log(a^b) = b*loga


I can't believe how much worse at Math I became since I went to university 4 years...


BTW, you seem to be pretty sharp in Math. Have you competed at the international level (IMOs and such)?

5^800 = 10^(800 log 5)

~ 10 ^ (800*.69897)

~ 10 ^ (559.17600)

= 10^559 * 10^.176

Which is a 560-digit number (10000....) times a 1-digit number, so you have a 560-digit number.

If you have a calculator handy, this is the easiest way.

Noisician's solution is far more beautiful and intelligent, however

quote:

Originally posted by Durafei LOL - my estimate was just "slightly" wrong For some reason I totally forgot that log(a^b) = b*loga I can't believe how much worse at Math I became since I went to university 4 years... BTW, you seem to be pretty sharp in Math. Have you competed at the international level (IMOs and such)?

Did you attend Waterloo University (if that's what it's called)? They own in the Putnam. I'd venture to guess Noisician has also had some math competition experience. True?

theres actually a really simple formula for calculating the number of digits in a number in any base b, and you can find that formula here:

http://mathworld.wolfram.com/Digit.html

and that formula gives approx 560 as the number of digits in 5^800.

it seems to me, tho, that noisician worked it out on the fly, so kudos to him for that

quote:

Originally posted by Noisician common logarithm can be extremely useful in such situation. with 5^800 i suppose u could do something like this log(5^800) = 800log5 = 800log(10/2) = 800(log10 - log2) = 800-800log2. now 800log2 = 80log(2^10) = 80log1024 > 80log1000 = 80*3 = 240 ie 800log2 > 240 also 80log1024-240 = 80(log1024-3) = 80(log1024-log1000) = 80log(1024/1000) = 80log(1000/1000+24/1000) = 80log(1+24/1000) = 80*24/1000*log(1+24/1000)^(1000/24) now since limn→0(1+n)^(1/n) = e < 3 80*24/1000*log(1+24/1000)^(1000/24) < 80*24/1000*log3 = 48/25*log3 < 2log3 = log9 < 1 ie 80log1024-240 < 1 <=> 80log1024 < 241 evaluate 240 < 80log1024 < 241 240 < 800log2 < 241 -241 < - 800log2 < -240 -241+800 < -800log2+800 < -240+800 559 < 800-800log2 < 560 559 < log(5^800) < 560 log(10^559) < log(5^800) < log(10^560) 10^559 < 5^800 < 10^560 your number has _exactly_ 560 digits. in general, u would need to use some arithmetic to estimate the value of logn first. once u've figured ou t the range, u should be able to get an answer. though it will be approximate in most cases...

Hehe, I knew you would be posting in this thread . Once again, thanks for your detailed explanation. Appreciate it .

Brain hurting... shutdown.

quote:

Originally posted by AnotherWay83 theres actually a really simple formula for calculating the number of digits in a number in any base b, and you can find that formula here: http://mathworld.wolfram.com/Digit.html and that formula gives approx 560 as the number of digits in 5^800. it seems to me, tho, that noisician worked it out on the fly, so kudos to him for that

So they do have a formula for that! Sweet .

quote:

Originally posted by Flyboy217 Did you attend Waterloo University (if that's what it's called)? They own in the Putnam. I'd venture to guess Noisician has also had some math competition experience. True?

I'm at UW now.. We used to own in Putnam - true, but not anymore. All the smart kids have graduated, and now all great students go to US schools.

quote:

Originally posted by AnotherWay83 theres actually a really simple formula for calculating the number of digits in a number in any base b, and you can find that formula here: http://mathworld.wolfram.com/Digit.html and that formula gives approx 560 as the number of digits in 5^800. it seems to me, tho, that noisician worked it out on the fly, so kudos to him for that

Indeed, that is a general formula for the solution. In the case of the question posted here, it simplifies a bit to the soln presented above.

The number of digits in m^n is

[n*log m] + 1

But I suggest reading and understanding Noisician's post in today's world over calculator over-reliance.