PYTHON SOLUTION

c = 0
for n in range(i,j+1):
    a = int(str(n)[::-1])
    if abs(n-a)%k == 0:
        c += 1 
return c           

def beautifulDays(i, j, k): count = 0 for r in range(i,j+1): str1 = str(r) rev_str = str1[::-1] rev_num = int(rev_str) diff = abs(r-rev_num) num = diff/k num1 = diff//k if num == num1: count += 1 return count

my very easy cpp solution

int beautifulDays(int i, int j, int k) {

int rem,count=0,temp;

for(int m=i;m <=j ; m++){///for looping
    temp=m;
    int rev=0;
    while(temp > 0){
    }

    if(abs(m - rev) % k ==0){  \\for checking it is equallly divisible or not
        count++;

    }
}
return count;

}

my very easy cpp solution

int beautifulDays(int i, int j, int k) {

int rem,count=0,temp;

for(int m=i;m <=j ; m++){///for looping
    temp=m;
    int rev=0;
    while(temp > 0){
         rem = temp % 10;
         rev = (rev * 10 ) + rem;
         temp = temp/10;  //for reversing number
    }

    if(abs(m - rev) % k ==0){  \\for checking it is equallly divisible or not
        count++;

    }
}
return count;

}

Here is my solution in JavaScript:

` let count = 0

for(let n = i; n <= j; n++){

const reversed = parseInt(n.toString().split('').reverse().join(''), 10);

const diff = Math.abs(n - reversed);

if(diff % k === 0){

    count++
}

}

return count; `