Is sum of divisors function a one-to-one function?

Suppose N N is a positive integer that can be expressed as a product of distinct prime numbers: N = p 1 × p 2 × × p n . N = p_1 \times p_2 \times \cdots \times p_n. Then the sum of all the positive divisors of N N is ( p 1 + 1 ) × ( p 2 + 1 ) × × ( p n + 1 ) . (p_1 + 1) \times (p_2 +1 ) \times \cdots \times (p_n + 1) . Now, suppose that M M is a positive integer the sum of whose positive divisors is ( 3 + 1 ) × ( 5 + 1 ) × ( 7 + 1 ) . (3 + 1) \times (5+1) \times (7 +1) . Does that mean M M must be equal to 3 × 5 × 7 ? 3\times5\times7?

Yes No

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1 solution

David Vreken
Mar 29, 2018

No, because M M could also equal 141 141 , since its divisors also add up to 1 + 3 + 47 + 141 = 192 = ( 3 + 1 ) × ( 5 + 1 ) × ( 7 + 1 ) 1 + 3 + 47 + 141 = 192 = (3 + 1) \times (5 + 1) \times (7 + 1) .

Or note that 191 191 is prime, so the sum of its divisors is also 192 192 .

Chris Lewis - 1 year, 7 months ago

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