Impossible Cryptarithm - (Part 4)

Computer Science Level 5

A B C ‾ × D E ‾ = F G H ‾ × I J ‾ = X \Large \overline{ABC} \times \overline{DE} = \overline{FGH} \times \overline{IJ} = X

If A , B , C , D , E , F , G , H , I , J A,B,C,D,E,F,G,H,I,J are distinct non-negative digits , find the minimum value of X X .

A B C ‾ , F G H ‾ \overline{ABC} , \overline{FGH} are three digit numbers , D E ‾ , I J ‾ \overline{DE} ,\overline{IJ} are two digit numbers.

This problem is a part of this set (click here) .


The answer is 8262.

View wiki
Nihar Mahajan by Nihar Mahajan , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Garrett Clarke
Aug 16, 2015

306 × 27 = 459 × 18 = 8262 306\times27 = 459\times18 = \boxed{8262}

Why is it minimal? Because I checked it with good, old-fashioned Java!

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Pranjal Jain
Nov 30, 2015 
1
2
3
4
5
6
7
 
from itertools import permutations
a=[]
for i in permutations('0123456789'):
    A,B,C,D,E,F,G,H,I,J=tuple(i)
    if '0' not in A+D+F+I and int(A+B+C)*int(D+E)==int(F+G+H)*int(I+J):
        a.append(int(A+B+C)*int(D+E))
print(min(a))

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...