Given a number N. The task is to find the largest good number in the divisors of given number N. A number X is defined as the good number if there is no such positive integer a > 1 such that a^2 is a divisor of X.
Input: N = 10 Output: 10 In 1, 2, 5, 10. 10 is the largest good number Input: N = 12 Output: 6 In 1, 2, 3, 4, 6, 12. 6 is the largest good number
Approach: Find all prime divisors of N. Assume they are p1, p2, …, pk (in O(sqrt(n)) time complexity). If the answer is a, then we know that for each 1 <= I <= k, obviously, a is not divisible by pi^2 (and all greater powers of pi). So a <= p1 × p2 ×… × pk. And we know that p1 × p2 × … × pk is itself a good number. So, the answer is p1 × p2 ×…× pk.
Below is the implementation of above approach:
- Sum of all second largest divisors after splitting a number into one or more parts
- Find sum of divisors of all the divisors of a natural number
- Find sum of inverse of the divisors when sum of divisors and the number is given
- Find the number of integers x in range (1,N) for which x and x+1 have same number of divisors
- Find the number of good permutations
- Find the minimum number of elements that should be removed to make an array good
- Find the sum of the number of divisors
- Find number from its divisors
- Find all divisors of a natural number | Set 2
- Find all divisors of a natural number | Set 1
- Find the number of divisors of all numbers in the range [1, n]
- Program to find count of numbers having odd number of divisors in given range
- Find largest sum of digits in all divisors of n
- Querying maximum number of divisors that a number in a given range has
- Check if a number is divisible by all prime divisors of another number
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.