# Answer to Question #69803 in Java | JSP | JSF for srevarun

Question #69803

Tyomitch calls the number with 2N digits (without leading zeroes) "interesting", if it's divisible

by both the number formed from its first N digits and the number formed from its last N digits.

For example, 1020 is "interesting" (divisible by 10 and 20) and 2005 is not. Tyomitch wants to

know how many "interesting" 2N-digit numbers exist. You are to help him.

Input

Input contains an integer N (1 ≤ N ≤ 10000).

Output

Output the number of "interesting" 2N-digit numbers.

by both the number formed from its first N digits and the number formed from its last N digits.

For example, 1020 is "interesting" (divisible by 10 and 20) and 2005 is not. Tyomitch wants to

know how many "interesting" 2N-digit numbers exist. You are to help him.

Input

Input contains an integer N (1 ≤ N ≤ 10000).

Output

Output the number of "interesting" 2N-digit numbers.

Expert's answer

public class InterestingNumbers {

public static void main(String[] args) {

int numbers [] = new int[100000];

int count = 0;

for (int i = 1; i <= numbers.length; i++) {

numbers[i - 1] = i;

if((Integer.toString(numbers[i - 1]).length() % 2) == 0) {

String numberToStr = Integer.toString(numbers[i - 1]);

String firstNToStr;

String lastToStr;

if (numberToStr.length() == 2) {

firstNToStr = numberToStr.substring(0, 1);

lastToStr = numberToStr.substring(1, 2);

}

else {

firstNToStr = numberToStr.substring(0, (numberToStr.length() / 2));

lastToStr = numberToStr.substring(numberToStr.length() / 2, numberToStr.length() - 1);

}

if ((firstNToStr.charAt(0) == '0') || (lastToStr.charAt(0) == '0'))

continue;

int firstN = Integer.parseInt(firstNToStr);

int lastN = Integer.parseInt(lastToStr);

if(((numbers[i - 1] % firstN) == 0) && ((numbers[i - 1] % lastN) == 0)) {

count++;

System.out.println(numbers[i - 1]);

}

}

}

System.out.println("==========\n" + count);

}

}

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