The following long division problem is not correct. In fact, none of the digits is what it should be.
Change each of the digits to a different digit so that the arithmetic is correct.
Note that there is no remainder, no leading zeroes, and like digits are not necessarily all changed to the same digit.
(In reply to
One More by np_rt)
That "12" must be in the 90's. And the divisor is 2033. Also, the "2" from 4321 must be 3 or 4. There is no way for 2022 be in the 90's.
That means that the range of the divisor is 2333, with the exception of 25.
Edited on December 3, 2004, 6:49 pm

Posted by np_rt
on 20041203 18:43:23 