Product of digits (Editorial)

To find the product of the digits, we first need to extract each digit of the number $n$ separately. Two integer arithmetic operations are used for this:

• the integer division $n/10$ discards the last digit of the number;

• the remainder $n\%10$ equals the last digit of the number.

Let $n$ be a three-digit number, $a$ its hundreds digit, $b$ its tens digit and $c$ its units digit, so that n = 100a + 10b + c. Then:

• the hundreds digit $a$ equals $n/100$: integer division by $100$ discards the last two digits;

• the tens digit $b$ equals $n/10\%10$: division by $10$ discards the units digit, and the tens digit becomes the last one;

• the units digit $c$ equals $n\%10$.

It remains to compute and print the product a \cdot b \cdot c.

Example

For n = 235:

• a = 235 / 100 = 2;

• b = 235 / 10~\%~10 = 23~\%~10 = 3;

• c = 235~\%~10 = 5.

The product of the digits is 2 \cdot 3 \cdot 5 = 30.

Algorithm implementation

Read the input data.

scanf("%d",&n);

Compute the hundreds digit $a$, tens digit $b$, and units digit $c$.

a = n / 100;
b = n / 10 % 10;
c = n % 10;

Compute and print the product of the digits.

res = a * b * c;
printf("%d\n",res);

Java implementation

import java.util.*;

public class Main
{
  public static void main(String []args)
  {
    Scanner con = new Scanner(System.in);
    int n = con.nextInt();
    int a = n / 100;
    int b = n / 10 % 10;
    int c = n % 10;
    int res = a * b * c;
    System.out.println(res);
    con.close();
  }
} 

Python implementation

Read the input data.

n = int(input())

Compute the hundreds digit $a$, tens digit $b$, and units digit $c$.

a = n // 100
b = n // 10 % 10
c = n % 10

Compute and print the product of the digits.

res = a * b * c
print(res)

Python implementation — digits

Read the input data.

n = input()

Compute and print the product of the digits.

res = int(n[0]) * int(n[1]) * int(n[2])
print(res)