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Place Ten Positive Integers, Get Minimum (Posted on 2023-04-25) Difficulty: 3 of 5
Consider a positive integer constant n.

Ten distinct integers are placed on the vertices of a regular decagon satisfying both the following conditions:
• The product of two non adjacent integers on the decagon is a multiple of n.
• The product of any pair of adjacent integers is NOT a multiple of n.

Determine the minimum value of n.

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

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Solution I think this is it | Comment 2 of 5 |
 n =
                6469693230
                
n is the product of the first ten primes. Each vertex's integer is this product divided by a unique pair of these prime numbers, one member of which is shared with each of its neighboring vertices, as in the following list:               
           
             Label    divisor  
 vertex # (quotient)   of n    
                
        1  111546435   29*2 
        2 1078282205    2*3 
        3  431312882    3*5 
        4  184848378    5*7 
        5   84021990    7*11
        6   45242610   11*13
        7   29274630   13*17
        8   20030010   17*19
        9   14804790   19*23
       10    9699690   23*29
       
For example, neither of the first two vertices is divisible by 2, so their product is also odd; but n is even, so it can't divide into that product.

111546435*1078282205 = 120278535891689175

120278535891689175/6469693230 = 8591323992263512/462120945
               =~ 18591072.5
               (better approx. is 18591072.49999999891803216147)
               
               
               
               
n=prod(primes(nthprime(10)))
for i=1:10
  prev=mod(i+8,10)+1;
  vertex=n/(nthprime(prev)*nthprime(i));
  fprintf('       %2d %10d   %2d*%-2d\n',i,vertex,nthprime(prev),nthprime(i))
end               

  Posted by Charlie on 2023-04-25 11:50:28
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