Simple Math Exercise – Computer V.S. Human

I came across this simple question: A entry level math question for kids,

A, B, C and D are 1-digit number from 0 to 9, find all possibilities for the following multiplication:

  A B C D
X       9
---------
  D C B A

If you are a programmer, you probably think that this is too easy, it can be solved almost instantly by brute-forcing all numbers, four nested loops with each number from 0 to 9 (the most naive solution) will be good. Alternatively, you can loop a single number from 0 to 9999 and check the result (multiplied by 9) to see if it is exactly the reverse (a bit more complicated).

The straightforward C++ implementation prints two answers: 0000 * 9 = 0000 and 1089 * 9 = 9801.

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// helloacm.com
#include <iostream>
 
using namespace std;
 
int main()
{
    for (int a = 0; a < 10; a ++) {
        for (int b = 0; b < 10; b ++) {
            for (int c = 0; c < 10; c ++) {
                for (int d = 0; d < 10; d ++) {
                    int n1 = a * 1000 + b * 100 + c * 10 + d;
                    int n2 = d * 1000 + c * 100 + b * 10 + a;
                    if (n1 * 9 == n2) {
                        cout << "a = " << a << endl;
                        cout << "b = " << b << endl;
                        cout << "c = " << c << endl;
                        cout << "d = " << d << endl;
                    }
                }
            }
        }
    }
    return 0;
}

// helloacm.com
#include <iostream>

using namespace std;

int main()
{
    for (int a = 0; a < 10; a ++) {
        for (int b = 0; b < 10; b ++) {
            for (int c = 0; c < 10; c ++) {
                for (int d = 0; d < 10; d ++) {
                    int n1 = a * 1000 + b * 100 + c * 10 + d;
                    int n2 = d * 1000 + c * 100 + b * 10 + a;
                    if (n1 * 9 == n2) {
                        cout << "a = " << a << endl;
                        cout << "b = " << b << endl;
                        cout << "c = " << c << endl;
                        cout << "d = " << d << endl;
                    }
                }
            }
        }
    }
    return 0;
}

Wait, you don’t need computer to find this at all! The 0000 is very obvious but how about the other solution?

A can only be 1 (if it is 2 or greater, D is two digit)
D then is 9
B then is 0 (the case of 1 can be excluded easily)
and C is 8

so the answer is 1089, faster than the computer.

–EOF (The Ultimate Computing & Technology Blog) —