All are 6-digit numbers with the same 6 digits

Computer Science Level 4

Given that A A is a 6-digit number and that 5 A 2 \dfrac{5A}{2} , 3 A 3A , 7 A 2 \dfrac{7A}{2} , 4 A 4A and 11 A 2 \dfrac{11A}{2} are all 6-digit numbers with the same 6 digits as A A . Find A A .


The answer is 153846.

Chan Lye Lee by Chan Lye Lee , 44, Singapore

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6 solutions

Julien Marcuse
Jun 5, 2018

My Java solution: 

import java.util.ArrayList;
public class Main {

     public static void main(String []args) {
        for (int i = 1; i <= 999999; i++) {
            ArrayList<Integer> digits = digits(i);
            if (!digits.containsAll(digits((5 * i) / 2))) {
                continue;
            }
            if (!digits.containsAll(digits(3 * i))) {
                continue;
            }
            if (!digits.containsAll(digits((7 * i) / 2))) {
                continue;
            }
            if (!digits.containsAll(digits((11 * i) / 2))) {
                continue;
            }
            if (!digits.containsAll(digits(4 * i))) {
                continue;
            }
            System.out.println(i);
        }
     }

     private static ArrayList<Integer> digits(int num) {
         ArrayList<Integer> digits = new ArrayList<>();
         while (num > 0) {
             digits.add(num % 10);
             num /= 10;
         }
         return digits;
     }
}
1 Helpful 0 Interesting 0 Brilliant 0 Confused

Is there a way to optimize?

Robert Tobio - 3 years ago

Am I the only one who thinks that a 6-digit number that satisfies all of this is cool?

Sanchit Sharma - 2 months, 1 week ago
Krutarth Patel
May 28, 2016 
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#include<iostream>
#include<algorithm>
#include<string>
using namespace std;

int main()
{
    for (int i = 100000; i <= 999999; i += 2)
    {
        string s, s1, s2, s3, s4;
        s = to_string(i), s1 = to_string((i << 1) + (i >> 1)), s2 = to_string((i << 1) + i), s3 = to_string(i << 2), s4 = to_string((i << 2) + i + (i >> 1));
        sort(s.begin(), s.end()), sort(s1.begin(), s1.end()), sort(s2.begin(), s2.end()), sort(s3.begin(), s3.end()), sort(s4.begin(), s4.end());
        if (s == s1 && s1 == s2 && s2 == s3 && s3 == s4)
        {
            cout << i << "\n";
            break;
        }
    }
    return 0;
}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

If this were to require more calculations, you could reduce the upper bound to 1,000,000*2/11.

First Last - 4 years, 6 months ago

I screened as I progressed. I realized that the initial number had to be between 100000 and 181818 and only even integers needed to be tested.

Do [ d1 = Sort [ IntegerDigits [ i ] ] ; j = 5 i 2 ; d2 = Sort [ IntegerDigits [ j ] ] ; If [ d1 = d2 , j = 3 i ; d2 = Sort [ IntegerDigits [ j ] ] ; If [ d1 = d2 , j = 7 i 2 ; d2 = Sort [ IntegerDigits [ j ] ] ; If [ d1 = d2 , j = 4 i ; d2 = Sort [ IntegerDigits [ j ] ] ; If [ d1 = d2 , j = 11 i 2 ; d2 = Sort [ IntegerDigits [ j ] ] ; If [ d1 = d2 , Print [ i ] ] ] ] ] ] , { i , 100000 , 181818 , 2 } ] ⟹ 153846 \text{Do}\left[\text{d1}=\text{Sort}[\text{IntegerDigits}[i]]; \\ j=\frac{5 i}{2};\text{d2}=\text{Sort}[\text{IntegerDigits}[j]]; \\ \text{If}\left[\text{d1}=\text{d2}, \\ j=3 i;\text{d2}=\text{Sort}[\text{IntegerDigits}[j]]; \\ \text{If}\left[\text{d1}=\text{d2}, \\ j=\frac{7 i}{2};\text{d2}=\text{Sort}[\text{IntegerDigits}[j]]; \\ \text{If}\left[\text{d1}=\text{d2}, \\ j=4 i;\text{d2}=\text{Sort}[\text{IntegerDigits}[j]]; \\ \text{If}\left[\text{d1}=\text{d2}, \\ j=\frac{11 i}{2};\text{d2}=\text{Sort}[\text{IntegerDigits}[j]]; \\ \text{If}[\text{d1}=\text{d2}, \\ \text{Print}[i]]\right]\right]\right]\right], \\ \{i,100000,181818,2\}\right] \Longrightarrow 153846 .

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Vlad Vasilescu
Jul 8, 2018

A not so brief Python solution :

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Is there any solution which does not use programming I tried my best but could not find any such solution.

Srikanth Tupurani - 2 years, 8 months ago
Eric Schneider
Jun 14, 2018

The range is bounded by the minimal 6-digit number (1000000) and the maximal number for which 11 A 2 \frac{11A}{2} is 6-digits (181818). A pythonic solution is implemented below. 

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print(sum([i if i%2==0and[sorted(str(x))for x in[i,5*i//2,3*i,7*i//2,4*i,11*i//2]].count(sorted(str(i)))==6else 0for i in range(100000,181819)]))

I love Python.

Running this program gives the result of 153846 \boxed{153846} .

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Zeeshan Ali
Jun 3, 2018 
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import itertools # https://docs.python.org/2/library/itertools.html#itertools.product
import numpy as np # http://www.numpy.org/

# set of all the possible 9 non-zero-digit numbers in acending order
product = list(itertools.product(range(10), repeat=6))

# exclude all those six-digit numbers starting with zero
product = [p for p in product if p[0]]

for p in product: # for every p = (a, b, c, d, e, f) in above set
    q = int(str(p[0])+str(p[1])+str(p[2])+str(p[3])+str(p[4])+str(p[5])) # A
    if (str(np.unique([int(x) for x in str(5*q//2)])) == str(np.unique([int(x) for x in str(3*q)]))
        and str(np.unique([int(x) for x in str(3*q)])) == str(np.unique([int(x) for x in str(7*q//2)]))
        and str(np.unique([int(x) for x in str(7*q//2)])) == str(np.unique([int(x) for x in str(4*q)]))
        and str(np.unique([int(x) for x in str(4*q)])) == str(np.unique([int(x) for x in str(11*q//2)]))):
        print (q) # print the elements or digits of that number 'p'
        break

153846

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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