Given a number, determine whether it is a valid Hyperperfect Number.
A number n is called k-hyperperfect if: n = 1 + k ∑idi where all di are the proper divisors of n.
Taking k = 1 will give us perfect numbers.
The first few k-hyperperfect numbers are 6, 21, 28, 301, 325, 496, 697, … with the corresponding values of k being 1, 2, 1, 6, 3, 1, 12, …
Input : N = 36, K = 1 Output : 34 is not 1-HyperPerfect Explanation: The Divisors of 36 are 2, 3, 4, 6, 9, 12, 18 the sum of the divisors is 54. For N = 36 to be 1-Hyperperfect, it would require 36 = 1 + 1(54), which we see, is invalid Input : N = 325, K = 3 Output : 325 is 3-HyperPerfect Explanation: We can use the first condition to evaluate this as K is odd and > 1 so here p = (3*k+1)/2 = 5, q = (3*k+4) = 13 p and q are both prime, so we compute p^2 * q = 5 ^ 2 * 13 = 325 Hence N is a valid HyperPerfect number
1570153 is 12-HyperPerfect 321 is not 3-HyperPerfect
Given k, we can perform a few checks in special cases to determine whether the number is hyperperfect:
- If K > 1 and K is odd , then let p = (3*k+1)/2 and q = 3*k+4 . If p and q are prime, then p2q is k-hyperperfect
- If p and q are distinct odd primes such that K(p + q ) = pq – 1 for some positive integral value of K, then pq is k-hyperperfect
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Improved By : jit_t