Unable to pass two test cases with this code

#1

#include
using namespace std;

int main(){

int n, armNo = 0, m;
int k;
cin >> m;
n = m;


while(n != 0){
     k = n%10;
    armNo = armNo + k*k*k;
    n = n/10;
}
if(m == armNo){
    cout << "true" << endl;
}
else{
    cout << "false" << endl;
}


return 0;

}

#2

Your code will give correct answer only for 3 digit numbers.
Say you are given a 4 digit number 1634
1634= 1^4 + 6^4 + 3^4 + 4^4 (So 1634 is an armstrong number)
Similarly:
153= 1^3 + 5^3 + 3^3 (So 153 is an armstrong number)
So by these examples you will get what an armstrong number is.
So to check if a number is armstrong number you just have to count the number of digits and extract each digit. Maintain a sum of (digit)^ noofdigits. If at the end of the loop, this sum==the given number then the number is armstrong else it is not armstrong.

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#3

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