Deficiency

Number Theory Level 2

How many deficient numbers are there?

Details and Assumptions

A positive integer n n is deficient iff the sum of all its positive divisors is strictly less than 2 n 2n .

many but not infinite Can't be determined 65 78 15 Infinitely Many

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Kalpok Guha by Kalpok Guha , 21, India

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2 solutions

Kalpok Guha
Apr 2, 2015

All primes are deficient numbers.

As σ ( p ) = p + 1 \sigma(p)=p+1 so it is deficient so all primes are deficient.

The answer is infinitely many.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Both constructions are nice! I personally prefer this one, but on the other hand Sophie's does not require proof of the infinitude of primes. Great problem.

Jake Lai - 6 years, 2 months ago

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Thank you Jake .It can be said all numbers in the form p k p^k are deficient numbers.

Kalpok Guha - 6 years, 2 months ago
Sophie Crane
Apr 3, 2015

If n = 2^k, sigma(n)= 2^(k+1) -1

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