Anyone solve Will Shortz's two week challenge (see details)?

You can see the challenge here It is about half way down the page under Next Week’s Challenge, which is really a two week challenge. It looked like it would be easy, but it has me stumped.

Here is how I approached it. I simplified things by turning the problem into an addition problem by moving the subtracted quantity to the other side. I also ignored the 2 digit move restriction. I just wanted to find a way of using the digits from 1 to 6 to get two numbers to add to a third.

We can eliminate the possibility of the sum being a 3 digit number. The smallest it could be is 123, which is out of range of any two digit plus one digit number.

We have ab + cd = ef. Assume no carry. a+c = e and b+d = f. a+c+e must be an even number and so must b+d+f. But that would mean the sum of the digits from 1 to 6 is even, which it is not because 1+2+3+4+5+6=21.

Suppose then that there is a carry. The only possibility is that b and d are 6 and 5 and f=1. That leaves 2, 3 and 4. The largest the sum digit e could be is 5, and 3+4=7 is already greater than 5 without the carry.

Shortz said there is no trick in the problem wording. What am I missing?