Write a program to find the four digit numbers, which are perfect squares, and all the digits in that number are even.
Answers
--You haven't told me in which program i have to write so i wrote in java,python java code can also work in c or c++ with changing the out statements
Program-in-java:
public class Main{
public static void main(String args[]){
int rem;
int[] arr = new int[4];
for(int i = 1000 ; i < 10000 ; i++){
for(int j = 32 ; j<101 ; j++ ){
if( i == (j*j)){
int k = 0,temp = i;
while(temp>0){
arr[k] = temp % 10;
temp = temp / 10;
k++;
}
if(arr[0]%2==0&&arr[1]%2==0&&arr[2]%2==0&&arr[3]%2==0){
System.out.print(i+"\t");
}
}
}
}
}
}
Output:
4624 6084 6400 8464
Program-in-python:
for i in range(1000,10000,1):
for j in range(32,100,1):
if i == j*j:
string = str(i)
if int(string[0])%2 == 0 and int(string[1])%2 == 0 and int(string[2])%2 == 0 and int(string[3])%2 == 0:
print(i)
Output:
4624
6084
6400
8464
Explanation:
- if you clearly see i used a outer for loop for getting all 4 digit numbers
- and i used inner loop to check which of the 3 digit number have 4 digit number as its square so if you can see i took 32 from starting because
- 31x31 = 961 which is not a 4 digit number
- 32x32 = 1024 so a perfect square of a four digit number will be above than 32
- and i took range 100 because 99x99 = 9801 which is last a 4 digit number have a perfect square.
- so i checked i with j*j and if that is true then i took each digit and checked if that is even or not if all the digits are even is true the i printed the number
- or else the loop iterates until all the numbers are printed and conditions breaks;
----Hope you liked my programs, if you liked it mark as brainliest, it would really help. :)
A program to find the four-digit numbers, which are perfect squares, and all the digits in that number are even:
- #include <iostream>
using namespace std;
bool isEven(int n){
int r,c=0;
while(n>0){
r = n%10;
if(r%2==0){
c++;
}
n = n/10;
}
if(c==4){
return true;
}
else{
return false;
}
}
bool isPerfectSquare(int n)
{
for (int i = 1; i * i <= n; i++) {
if ((n % i == 0) && (n / i == i)) {
return true;
}
}
return false;
}
int main()
{
int i;
for(i=1000; i<10000; i++){
if(isPerfectSquare(i) && isEven(i)){
cout<<i<<endl;
}
}
return 0;
}
What are perfect squares?
- Perfect squares are numbers that give a result of the product when we perform multiplication to itself that number. These numbers or integers can be negative, positive, or zero.
- The perfect square of 0 is 0, the perfect square of 1 is 1, the perfect square of 2 is 4. This we get by multiplying the number with itself like 0×0=0, 1×1=1, 2×2=4.