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.

This one concerns the 2 in the 4321. The 3 in the 365 is a 1 for sure. So that 12 must be in the 90's and the 6 in the 365 is a 0 as well meaning the 8 in 95867 is a 0 as well.

My last observation said that that the divisor is between 20 and 49. In order for the product of "2" times the divisor to be in the 90's, that means that the 2 must be a 3 or a 4.

Also that 1 in 4321 times the divisor is still 2 digits so it must be less than 5 (since the divisor is at least 20), meaning it's 2, 3 or 4.

This further narrows down the divisor to 20-33. And the quotients looks is wxyz. w is 1, 2, or 3. x is 0, y is 3 or 4. z is 2, 3, or 4.

This narrows it down to 18 values for the quotient and 14 values for the divisor.

Edited on December 3, 2004, 6:38 pm

Posted by np_rt on 2004-12-03 18:37:10