The fold is really creative but I'd be afraid of an off by one error. I could reason it out, but being sure they meet where I think they do at the center and th

The fold is really creative but I'd be afraid of an off by one error. I could reason it out, but being sure they meet where I think they do at the center and that I indeed can multiply by 50 would take more time than it should to convince myself. I'd instead look at trying to collect/aggregate the ones, tens, & hundreds column. It turns out this is kind of a symmetric problem. First, aggregate the singles column of this addition. You'll have 0-9, repeated 10 times. Add 1-9 and you get 45. Multiple by 10 since they're repeated for the aughts, the 10's, the 20's, &c. 450 is the total value of the first column of digits. Second, aggregate the ten's column. The order is different but the values are the same. so it'll come to the same 450. 450 * 10 = 4500. Last, add the hundreds column; just 100. 450 (singles aggregate) + 4500 (tens aggregate) + 100 (hundreds aggregate) = 5050. I really like the author's late conclusion on the many approaches to solving the problem: > What I discovered when I tried this experiment is that it's really hard to do it the hard way. You may set out to plod dutifully through all the addition operations, but shortcuts present themselves even when you're not looking for them.