Find the smallest positive integer such that there exists a positive integer whose sum of divisors is equal to .
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Since 7 , 1 2 4 9 are prime and therefore σ 1 ( 8 7 4 3 ) = σ 1 ( 7 ) σ 1 ( 1 2 4 9 ) = 8 × 1 2 5 0 = 1 0 0 0 0 , we see that k ≤ 4 . It is simple enough to check the values of σ 1 ( n ) for 1 ≤ n ≤ 1 0 0 0 to see that it is not possible for σ 1 ( n ) to be equal to 1 0 , 1 0 0 or 1 0 0 0 . Thus k = 4 .
For interest, the other number n with σ 1 ( n ) = 1 0 0 0 0 is 9 4 8 1 = 1 9 × 4 9 9 .