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Math Question Again

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!

Re: Math Question Again

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.

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.

quote:

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

well kinda... in 2001 i managed to reach the semi-finals of the national mathematical olympiad. there were only six problems to solve, but at that time i was only able to solve three of them. two other problems i figured out within the next week at home, while the last (6th) problem took me more than a whole year to get it done. and of course it was stereometry-related; that's what they always use for the last and hardest problem of the given ones. oh and here it is, in case anyone cares:

given a sphere of radius r, find the locus of points through which it is possible to draw three tangent lines to the sphere that form a trihedral angle with three right faces.

quote:

Originally posted by Flyboy217 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

your solution is far better then Noisician's soluion which was just an over-complication of yours

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