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Here we will see, how to check, whether a number is full prime or not. A number is said to be a full prime, if it is prime, and all of its digits are also prime. Suppose a number is 37, this is full prime. But 97 is not full prime as 9 is not a prime number.One efficient approach is that; first we have to check whether any digit is present that is not prime. Digits must be in 0 to 9. In that range 2, 3, 5 and 7 are prime, others are not prime. If all are ... Read More
Here we will see, whether a number is power of another number or not. Suppose a number 125, and another number 5 is given. So it will return true when it finds that 125 is power of 5. In this case it is true. 125 = 53.AlgorithmisRepresentPower(x, y): Begin if x = 1, then if y = 1, return true, otherwise false pow := 1 while pow < y, do pow := pow * x done if pow = y, then return true return false ... Read More
Here we will see how to check whether a number is Mystery number or not. A Mystery number is a number that can be represented by sum of two numbers, and the numbers re reverse of each other. Let us see the code to get better idea. We have to check all pairs and find the decision.Example Live Demo#include using namespace std; int revNum(int str) { string s = to_string(str); reverse(s.begin(), s.end()); stringstream ss(s); int rev = 0; ss >> rev; return rev; } bool isMysteryNumber(int n) { for (int i=1; i
Here we will see, if a number has adjacent set bits in its binary representation. Suppose the number 12 has two consecutive 1s (12 = 1100).To check this type of number, the idea is very simple. We will shift the number 1 bit, then do bitwise AND. If the bitwise AND result is non-zero, then there must be some consecutive 1s.Example Live Demo#include using namespace std; bool hasConsecutiveOnes(int n) { if((n & (n >> 1)) == 1){ return true; }else{ return false; } } int main() { int num = 67; //1000011 if(hasConsecutiveOnes(num)){ cout
In this section we will check whether a number has same number of set bits and unset bits or not. Suppose the number 12 is there. Its binary representation is 1100. This has same number of 0s and 1s.The approach is simple. We will check each bit of the number, and if that is 1, then increase set_bit_count, and if it is 0, then increase unset_bit_count. Finally, if they are same, then return true, otherwise false.Example Live Demo#include using namespace std; bool hasSameSetUnset(int n) { int set_count = 0, unset_count = 0; while(n){ if((n & ... Read More
In this section we will see, whether a number can be represented as tree consecutive numbers or not. Suppose a number is 27. This can be represented as 8 + 9 + 10.This can be solved in two different approach. The first approach is Naïve approach. In that approach, we have to check i + (i + 1) + (i + 2) is same as number or not. Another efficient approach is by checking whether the number is divisible by 3 or not. Suppose a number x can be represented by three consecutive 1s, then x = (y - 1) ... Read More
In this section, we will see if we can express one number as the sum of two triangular numbers or not. The triangular numbers are like below −From the example, we can see that 1, 3, 6, 10 are some triangular numbers. We need to express a number N (say 16) as sum of two triangular numbers (6, 10).The approach is very simple. We have to get all triangular numbers less than N. Form a set from these values. Now we have to take a number say X from the set, and check whether N – X is present in ... Read More
Here we will check whether we can represent a number as power, like xy or not. Suppose a number 125 is present. This can be represented as 53. Another number 91 cannot be represented as power of some integer value.AlgorithmisRepresentPower(num): Begin if num = 1, then return true for i := 2, i2
Here we will check whether we can represent a number as power, like ab or not. Suppose a number 125 is present. This can be represented as 53. Another number 91 cannot be represented as power of some integer value.AlgorithmisRepresentPower(num): Begin if num = 1, then return true for i := 2, i2
Here we will check whether we can represent a number like ab or not. Suppose a number 125 is present. This can be represented as 53. Another number 91 cannot be represented as power of some integer value.AlgorithmisRepresentPower(num): Begin if num = 1, then return true for i := 2, i2