What are the possible values of a six digit positive integer N, where all digits of N are different, and 4*N has the same six digits as N?
Prove that there are no others.
Roughly how many numbers need to be considered?
Well, first two digits must be between 12 and 24, but not 22, which is 12 possible combinations.
The third one can be any of 8 remaining digits.
The fourth one can be any of 7 remaining digits.
The fifth one can be any of 6 remaining digits.
The last one can be any of 5 remaining digits.
So at first glance we have 12*8*7*6*5 = 20,160 possibilities.
But the number must equal 0 mod 9, because N and 4N have the same digits.
And approximately 1/9 of the 20,160 possibilities meet this criteria.
So I guess that there are only 2,240 viable candidates for N.
I can think of no analytic way to narrow these down. Time for somebody to write a computer program!
Let me count the ways
