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 Integer Illation From Divisors (Posted on 2016-01-27)
Find all possible values of a positive integer M such that:
M has precisely 16 positive integer divisors D(1), D(2), ......, D(16), and:
1 = D(1) < D(2) < ......< D(16) = M, and:
D(6) = 18, D(9) – D(8) = 17

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution | Comment 1 of 2
D(6)=18 means that 1, 2, 3, 6, 9, 18 are all divisors of M.  That accounts for all of D(1)-D(6).  So M=2^x*3^y*n^z for some integers x>=1, y>=2, and z>=0.

x must be 1, otherwise 4 would be a divisor which contradicts the requirement D(6)=18.

z=0 or 1, otherwise there would be at least 18 divisors which contradicts the requirement M has precisely 16 positive integer divisors.

If z=0 then y=7 and M=2*3^7.  D(8) and D(9) then would be 54 and 81.  81-54=27 so this is not a number M.

Then z=1.  y must be 3 for M to have 16 divisors, then M=2*3^3*n.  The 8 known divisors are then 1, 2, 3, 6, 9, 18, 27, 54.  This also implies n must be prime.

If n>54 then D(9)=n and D(8)=54.  n-54=17 has a solution of n=71.  71 is prime so forms a solution M = 2*3^3*71 = 3834.

If 54>n>27 then D(9)=54 and D(8)=n.  54-n=17 has a solution of n=37.  37 is prime so forms a solution M = 2*3^3*37 = 1998.

If 27>n then D(9)=54 and D(8)=27.  54-27=27, not 17 so cannot form a solution.

In summary, M = 3834 or 1998.

 Posted by Brian Smith on 2016-01-27 10:02:01

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