2(p+q) >= pq
if p > 2, then q <= (2p+2)/(p-2)
It is easy to show that if p>11 there is no prime q which satisfies the condition
So, p is either 3,5,7
and you can find the values of q which will work
The proper factors of a number n are the factors which are less than n. A number n is deficient if the sum of its proper factors is less than n.
For example, 22 is deficient since 1 + 2 + 11 = 14 < 22.
A number n is super-deficient if twice the sum of its proper factors is less than n. For example, all odd primes p are super-deficient since 2 × 1 = 2 < p.
A number n = pq, where p and q are different odd primes, is not super-deficient. Find all possible values for n. Show that there are no more.
1 answer