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# Absolute difference between the first X and last X Digits of N?

Here we will see how to get the differences between first and last X digits of a number N. The number and X are given. To solve this problem, we have to find the length of the number, then cut the last x digits using modulus operator. After that cut all digits from the number except first x digits. Then get the difference, and return the result. Let the number is N = 568424. The X is 2 so first two digits are 56, and last two digits are 24. The difference is (56 - 24) = 32.

## Algorithm

#### diffFirstLastDigits(N, X)

begin p := 10^X last := N mod p len := length of the number N while len is not same as X, do N := N / 10 len := len -1 done first := len return |first - last| end

## Example

#include <iostream> #include <cmath> using namespace std; int lengthCount(int n){ return floor(log10(n) + 1); } int diffFirstLastDigits(int n, int x) { int first, last, p, len; p = pow(10, x); last = n % p; len = lengthCount(n); while(len != x){ n /= 10; len--; } first = n; return abs(first - last); } main() { int n, x; cout << "Enter number and number of digits from first and last: "; cin >> n >> x; cout << "Difference: " << diffFirstLastDigits(n,x); }

## Output

Enter number and number of digits from first and last: 568424 2 Difference: 32

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