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Minimum Value Muse 3 (Posted on 2023-11-12) Difficulty: 3 of 5
Determine the minimum value of a positive integer N such that each of N, N+1, N+2, N+3, N+4, and N+5 has the same number of divisors (including 1 and itself.)

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Solution computer solution Comment 2 of 2 |
d=0; ct=0;
for n=1:9999999
  prev=d;
  d=length(divisors(n));
  if d==prev
    ct=ct+1;
  else
    ct=1;
  end
  if ct>=6
    disp([n-ct+1 n ct])
  end
end

finds

 28374 through 28379 have the same number of divisors.
 
Each has 8 divisors as seen by

>> for i=28374:28379 fprintf('%d ',divisors(i)), fprintf('\n')
end
1 2 3 6 4729 9458 14187 28374 
1 5 25 125 227 1135 5675 28375 
1 2 4 8 3547 7094 14188 28376 
1 3 9 27 1051 3153 9459 28377 
1 2 7 14 2027 4054 14189 28378 
1 13 37 59 481 767 2183 28379 


 
 

  Posted by Charlie on 2023-11-12 08:56:56
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