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2^(10n) (Posted on 2015-04-15) Difficulty: 2 of 5
210 = 1024
220 = 1048576
Note that raising 2 to each of the first two multiples of 10 results in a number whose first digit is 1.

Find the smallest multiple of 10 where 2 raising to that power results in a number that does not begin with 1.

See The Solution Submitted by Jer    
Rating: 1.0000 (1 votes)

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re(4): worked out Comment 8 of 8 |
(In reply to re(3): worked out by Ady TZIDON)

I still do not understand your comment.


1024^1 = 2^10 begins with a 1; 1024^2 = 2^20 begins with a 1; etc.

A table of powers of 1024 = powers of 2^10 = powers of 2^(10k):

  1                      1024                                                   
  2                      1048576                                                
  3                      1073741824                                             
  4                      1099511627776                                          
  5                      1125899906842625                                       
  6                      1.152921504606845D+18                                   7                      1.180591620717406D+21                                   8                      1.20892581961463D+24                                  
  9                      1.237940039285386D+27                                   10                     1.267650600228232D+30 
  21                     1.645504557321195D+63                               
  22                     1.684996666696936D+66                               
  23                     1.725436586697645D+69                                 24                     1.766847064778371D+72                                 25                     1.809251394333035D+75                                 26                     1.85267342779701D+78                                   27                     1.897137590064173D+81                               
  28                     1.942668892225694D+84                                 29                     1.989292945639148D+87                              

At k=29, the powers of 2^(10k) still start with 1 and it is the last such.

At k = 30 this suddenly changes. Now they start with 2:

  30                     2.037035976334467D+90                               
  31                     2.085924839766534D+93                                 32                     2.13598703592091D+96                                   33                     2.18725072478299D+99                                   34                     2.239744742177823D+102                               35                     2.29349861599007D+105                                 36                     2.348542582773874D+108                               37                     2.404907604760423D+111                               38                     2.46262538727472D+114                                 39                     2.521728396569217D+117                                 
I don't disagree that there were numerous values that begin with 1; that was my point, and Jer's point in writing the puzzle.

BTW, you say " The 30 is the 1st number  m,  for which 2^(10m)begins with the digit 1." Actually it's the first that begins with 2, also my point.

What is the point of checking 2^291 through 2^299; those exponents are not multiples of 10. They are examples of numerous powers that both do and do not begin with 1. There are numerous examples of 2^k, lower than 2^300 that do not begin with 1, if that was your point, but none of those k were multiples of 10. The problem was really to find the power of 1024 that does not begin with 1, and that is what I was looking for, and that was the use of the word still.





  Posted by Charlie on 2015-04-17 14:09:41
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