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Topic: A Beastly Number (Read 1806 times) 

ThudnBlunder
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A Beastly Number
« on: Nov 26^{th}, 2008, 9:05am » 
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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 » 
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Barukh
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Re: A Beastly Number
« Reply #1 on: Dec 2^{nd}, 2008, 10:32am » 
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For an easier part b), I get: 2^{664}(5^{667}  1)


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Barukh
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Re: A Beastly Number
« Reply #2 on: Dec 3^{rd}, 2008, 12:31am » 
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According to the following article, solving part a) may require calculation of a certain logarithm to a precision of more than 670 decimal digits!


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


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


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SMQ
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Re: A Beastly Number
« Reply #5 on: Dec 5^{th}, 2008, 12:32pm » 
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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 » 
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Barukh
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Re: A Beastly Number
« Reply #6 on: Dec 6^{th}, 2008, 9:52am » 
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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.


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


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Barukh
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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.


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Barukh
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Re: A Beastly Number
« Reply #9 on: Dec 7^{th}, 2008, 11:21pm » 
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I get the following answer to part a): 13407273847...


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Eigenray
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Re: A Beastly Number
« Reply #10 on: Dec 8^{th}, 2008, 7:57am » 
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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 » 
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Barukh
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Re: A Beastly Number
« Reply #11 on: Dec 8^{th}, 2008, 10:19am » 
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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?


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Eigenray
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Re: A Beastly Number
« Reply #12 on: Dec 8^{th}, 2008, 12:44pm » 
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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 » 
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ThudnBlunder
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Re: A Beastly Number
« Reply #13 on: Dec 9^{th}, 2008, 4:03pm » 
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My source for this problem got yet a different answer, 1.340727397...

« Last Edit: Dec 9^{th}, 2008, 4:06pm by ThudnBlunder » 
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Eigenray
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Re: A Beastly Number
« Reply #14 on: Dec 9^{th}, 2008, 5:32pm » 
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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.


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