("47x"*729*345*343)
I assume you mean "47x" to be a three-digit number where
4 is the hundreds digit, 7 is the tens digit and x is
the units digit. It cannot mean 47 times x, otherwise
there would be infinitely many answers.
The result will have the same last digit as the last digit
of the product of the last digits:
x*9*5*3 = 135*x
That will have the last digit the same as the last digit of
5*x, and
5*1=5, 5*2=10, 5*3=15, 5*4=20, 5*5=25, 5*6=30, 5*7=35, 5*8=40, 5*9=45
So we see that if x is an even digit, the last digit will be 0, and
if x is an odd digit, the last digit will be 5.
Therefore x must be an odd digit, either 1,3,5,7, or 9
Answer: The maximum number of digits x can be is 5.
Edwin
