First few Blum Integers are 21, 33, 57, 69, 77, 93, 129, 133, 141, 161, 177, …
Note: Because of the condition that both the factors should be semi-primes, even numbers can not be Blum integers neither can be the numbers below 20,
So we have to check only for an odd integer greater than 20 that if it is a Blum Integer or not.
Input: 33 Output: Yes Explanation: 33 = 3 * 11, 3 and 11 are both semi-primes as well as of the form 4t + 3 for t = 0, 2 Input: 77 Output: Yes Explanation: 77 = 7 * 11, 7 and 11 both are semi-prime as well as of the form 4t + 3 for t = 1, 2 Input: 25 Output: No Explanation: 25 = 5*5, 5 and 5 are both semi-prime but are not of the form 4t + 3, Hence 25 is not a Blum integer.
Approach: For a given odd integer greater than 20, we calculate the prime numbers from 1 to that odd integer. If we find any prime number that divides that odd integer and its quotient both are prime and follow the form 4t + 3 for some integer, then the given odd integer is Blum Integer.
Below is the implementation of above approach :
- Find One's Complement of an Integer
- Minimum positive integer value possible of X for given A and B in X = P*A + Q*B
- Find whether a given integer is a power of 3 or not
- Convert given integer X to the form 2^N - 1
- Greatest Integer Function
- Smallest integer which has n factors or more
- Replace all ‘0’ with ‘5’ in an input Integer
- Square root of an integer
- Convert a String to an Integer using Recursion
- Check if the given array contains all the divisors of some integer
- Print first k digits of 1/n where n is a positive integer
- Smallest integer with digit sum M and multiple of N
- Check for integer overflow on multiplication
- Generate all unique partitions of an integer
- Generate integer from 1 to 7 with equal probability
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