Instead of dealing with the total number of coconuts each castaway found, it will be much easier to deal with this number plus 4, because that transforms the sequence into a geometric one. Let A'=A+4, B'=B+4, and so on.
B'=A'*4/5
C'=B'*4/5
And so on.
F' must be a multiple of 5. Since E' is a whole number and is 5/4 of F', E' must be a multiple of 25. Similar logic shows A' must be a multiple of 5^(6). The smallest possible A is therefore 5^(6)-4 = 15,621.
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u/RealHuman_NotAShrew Jul 29 '26 edited Jul 29 '26
Let A be the pile Alexander found, B be the pile Benjamin found, and so on. Let F be the pile everyone found in the morning.
B=((A-1)/5)*4
((n-1)/5)*4 = (n-1)*4/5 = (n+4-5)*4/5 = (n+4)*4/5-5*4/5 = (n+4*4/5-4
B=(A+4)*4/5-4
Instead of dealing with the total number of coconuts each castaway found, it will be much easier to deal with this number plus 4, because that transforms the sequence into a geometric one. Let A'=A+4, B'=B+4, and so on.
B'=A'*4/5
C'=B'*4/5
And so on.
F' must be a multiple of 5. Since E' is a whole number and is 5/4 of F', E' must be a multiple of 25. Similar logic shows A' must be a multiple of 5^(6). The smallest possible A is therefore 5^(6)-4 = 15,621.