Author 
Topic: A Beastly Number (Read 1793 times) 

ThudnBlunder
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489


A Beastly Number
« on: Nov 26^{th}, 2008, 9:05am » 
Quote Modify

a) Find the first 6 digits of (10^{666})! b) How many trailing zeros does the above number have?

« Last Edit: Nov 26^{th}, 2008, 1:12pm by ThudnBlunder » 
IP Logged 
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.



Barukh
Uberpuzzler
Gender:
Posts: 2276


Re: A Beastly Number
« Reply #1 on: Dec 2^{nd}, 2008, 10:32am » 
Quote Modify

For an easier part b), I get: 2^{664}(5^{667}  1)


IP Logged 



Barukh
Uberpuzzler
Gender:
Posts: 2276


Re: A Beastly Number
« Reply #2 on: Dec 3^{rd}, 2008, 12:31am » 
Quote Modify

According to the following article, solving part a) may require calculation of a certain logarithm to a precision of more than 670 decimal digits!


IP Logged 



ThudnBlunder
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489


Re: A Beastly Number
« Reply #3 on: Dec 4^{th}, 2008, 6:56pm » 
Quote Modify

on Dec 2^{nd}, 2008, 10:32am, Barukh wrote:For an easier part b), I get: 2^{664}(5^{667}  1) 
 If we take n/4 as an estimate for the number of trailing zeros, we get 2.5*10^{665} Your number is 5 times this.


IP Logged 
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.



Barukh
Uberpuzzler
Gender:
Posts: 2276


Re: A Beastly Number
« Reply #4 on: Dec 5^{th}, 2008, 12:15pm » 
Quote Modify

on Dec 4^{th}, 2008, 6:56pm, ThudanBlunder wrote: Your number is 5 times this. 
 Yes, I should've written 5^{666} instead. But now I realized the answer is incorrect anyway (doesn't take into account fractions). To write the answer in a "compact form" may be as difficult as part a) then...


IP Logged 



SMQ
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 2084


Re: A Beastly Number
« Reply #5 on: Dec 5^{th}, 2008, 12:32pm » 
Quote Modify

2) _{}10^{666}/5^{n}_{} ^{n=1} Where in practice the upper bound can be reduced to _{}666 log 10 / log 5_{} = 952 Edit:  = 2^{666}_{}5^{666} / 4_{} +  _{286} ^{n=1}  _{}2^{666}/5^{n}_{}  Edit2: all of which I think I can safely assume anyone hanging around in Putnam already knew (and is clearly the driver for Barukh's answer), but since this thread has only seen a little action I thought I'd put it out there anyway. SMQ

« Last Edit: Dec 5^{th}, 2008, 1:19pm by SMQ » 
IP Logged 
SMQ



Barukh
Uberpuzzler
Gender:
Posts: 2276


Re: A Beastly Number
« Reply #6 on: Dec 6^{th}, 2008, 9:52am » 
Quote Modify

SMQ, you are right. My answer is wrong, since it doesn't take into account the second term in your formula  the sum which is challenging to evaluate. After working it out, I get the following answer to b): 2^{664}5^{666}  143. I will supply details later, after I find the answer to the first question.


IP Logged 



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: A Beastly Number
« Reply #7 on: Dec 6^{th}, 2008, 12:25pm » 
Quote Modify

on Dec 6^{th}, 2008, 9:52am, Barukh wrote:After working it out, I get the following answer to b): 2^{664}5^{666}  143. 
 So is there a clever way to compute the sum of the digits of 2^{666} = 34004...23324_{5}? Actually, the approximation 333*log_{5}(2) is pretty good here.


IP Logged 



Barukh
Uberpuzzler
Gender:
Posts: 2276

on Dec 6^{th}, 2008, 12:25pm, Eigenray wrote: So is there a clever way to compute the sum of the digits of 2^{666} = 34004...23324_{5}? 
 I don't know. I used highprecision software (MPFR) to compute the number (I still want to get a confirmation it's correct). Quote:Actually, the approximation 333*log_{5}(2) is pretty good here. 
 Yes, but it may be quite inaccurate for other exponents. In the attached graph, I plotted the discrepancies between your approximation and actual number for all cases 10^{n} with 10 < n < 2000. Again, if my calculations are correct.


IP Logged 



Barukh
Uberpuzzler
Gender:
Posts: 2276


Re: A Beastly Number
« Reply #9 on: Dec 7^{th}, 2008, 11:21pm » 
Quote Modify

I get the following answer to part a): 13407273847...


IP Logged 



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: A Beastly Number
« Reply #10 on: Dec 8^{th}, 2008, 7:57am » 
Quote Modify

on Dec 7^{th}, 2008, 11:21pm, Barukh wrote:I get the following answer to part a): 13407273847... 
 Are you sure it's not 134072738469787? But the first 6 digits are correct And the 143 is correct, as the following short (but rather inefficient) program shows: Code: #include<stdio.h> int A[300]; int main() { int n,i,c; A[0]=1; for(n=0;n<666;n++) for(c=i=0;i<300;i++) { c+=2*A[i]; A[i]=c%5; c/=5; } for(c=i=0;i<300;i++) c+=A[i]; printf("%i\n",c/4); } 


« Last Edit: Dec 8^{th}, 2008, 8:21am by Eigenray » 
IP Logged 



Barukh
Uberpuzzler
Gender:
Posts: 2276


Re: A Beastly Number
« Reply #11 on: Dec 8^{th}, 2008, 10:19am » 
Quote Modify

on Dec 8^{th}, 2008, 7:57am, Eigenray wrote: Are you sure it's not 134072738469787? 
 Hmm... I did calculate the logarithm with very high precision, so at least 15 digits should be accurate. Then, I did exponentiation with a double precision. Could it be I lost 5 digits of accuracy there? What's your method?


IP Logged 



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: A Beastly Number
« Reply #12 on: Dec 8^{th}, 2008, 12:44pm » 
Quote Modify

Code: In[1]:= u=1FractionalPart[ N[10^666 Log[10,E], 686] ] Out[1]= 0.72825054586682085824 In[2]:= v=N[Log[10,Sqrt[2 Pi]], 20] Out[2]= 0.39908993417905752478 In[3]:= 10^(u+v) Out[3]= 13.407273846978712508 
 What do you get? Edit: And here it is with MPFR: Code:#include<stdlib.h> #include<stdio.h> #include<gmp.h> #include<mpfr.h> int main() { mpfr_t u,v; mpfr_set_default_prec(10000); mpfr_init_set_ui(v,10,GMP_RNDN); mpfr_pow_ui(v,v,666,GMP_RNDN); //v=10^666 mpfr_init_set_ui(u,1,GMP_RNDN); mpfr_exp(u,u,GMP_RNDN); mpfr_log10(u,u,GMP_RNDN); //u=log_10 e mpfr_mul(u,u,v,GMP_RNDN); mpfr_frac(u,u,GMP_RNDN); //u=frac[10^666 log_10 e] mpfr_set_si(v,1,GMP_RNDN); mpfr_acos(v,v,GMP_RNDN); mpfr_mul_ui(v,v,2,GMP_RNDN); //v=2pi mpfr_log10(v,v,GMP_RNDN); mpfr_div_ui(v,v,2,GMP_RNDN); //v=log(sqrt(2pi)) mpfr_sub(v,v,u,GMP_RNDN); mpfr_ui_pow(v,10,v,GMP_RNDN); //10^(vu) mpfr_out_str(stdout,10,20,v,GMP_RNDN); printf("\n"); } 
 Gives 1.3407273846978712508 (I already had GMP but apparently it doesn't do logs. So I downloaded the MPFR sources and started ./configure; make. Then I realized I could install it through Cygwin, and wrote the above before it finished compiling.)

« Last Edit: Dec 8^{th}, 2008, 2:00pm by Eigenray » 
IP Logged 



ThudnBlunder
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489


Re: A Beastly Number
« Reply #13 on: Dec 9^{th}, 2008, 4:03pm » 
Quote Modify

My source for this problem got yet a different answer, 1.340727397...

« Last Edit: Dec 9^{th}, 2008, 4:06pm by ThudnBlunder » 
IP Logged 
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: A Beastly Number
« Reply #14 on: Dec 9^{th}, 2008, 5:32pm » 
Quote Modify

The funny thing is that he did compute the fractional part of 10^{666}/log(10) correctly to 14 digits. But he decided to round it to 8 digits before exponentiating, and then claim 10 digits of accuracy in the result: Quote:{2} x 10^{0.27174945} = 1.340727397, 
 which is silly.


IP Logged 



