Pow #1- a Digital Proof
A Digital Proof Write-up
The problem in A Digital Proof has two parts. The first is to fill in five boxes with numbers that fit the criteria: each box has a number, and the digit that is placed in each box must be the amount of times that number appears in the whole five digit number formed by the boxes. The second part of the problem is to prove that there is only one solution.
How I went about solving this problem was somewhat simple; at least, it was at first. I started from the ‘four’ box (the fifth box, labeled with a four). I realized that four wouldn’t work in that box, because that would mean that there were four fours, and that wouldn’t work. I couldn’t put three in the box, either, because that would require there to be three fours, and that wouldn’t work out either. Two didn’t work for the same reasons as four and three, and even one wasn’t a possibility. This left me with one option: zero.
One box down, four to go. Easy, right? That’s what I thought as I filled in the ‘three’ box, again with a zero for the same reasons that I’d put a zero in the ‘four’ box. Four wouldn’t work because that would require three to be in four boxes, and then that wouldn’t leave room for any other numbers. Again, this was the reason that three, two, and one didn’t work. For three, too, the only possibility was zero.
Up until now, things had been fairly straightforward. Then, once I hit the ‘two’ box, things began to get more complicated. Here, I couldn’t put four or three because two of the boxes had already been filled, and I couldn’t change that. Then, I tried two. This could work, but only if there was a two elsewhere. I couldn’t put a two in the ‘one’ box, but I could put it in the zero box, because of the ‘four’ and ‘three’ boxes. Good thing I didn’t change those. That left me with the ‘one’ box. There was really only one option for that box, and that was putting a one in it.
That was my process for solving the seemingly daunting, but surprisingly easy
The problem in A Digital Proof has two parts. The first is to fill in five boxes with numbers that fit the criteria: each box has a number, and the digit that is placed in each box must be the amount of times that number appears in the whole five digit number formed by the boxes. The second part of the problem is to prove that there is only one solution.
How I went about solving this problem was somewhat simple; at least, it was at first. I started from the ‘four’ box (the fifth box, labeled with a four). I realized that four wouldn’t work in that box, because that would mean that there were four fours, and that wouldn’t work. I couldn’t put three in the box, either, because that would require there to be three fours, and that wouldn’t work out either. Two didn’t work for the same reasons as four and three, and even one wasn’t a possibility. This left me with one option: zero.
One box down, four to go. Easy, right? That’s what I thought as I filled in the ‘three’ box, again with a zero for the same reasons that I’d put a zero in the ‘four’ box. Four wouldn’t work because that would require three to be in four boxes, and then that wouldn’t leave room for any other numbers. Again, this was the reason that three, two, and one didn’t work. For three, too, the only possibility was zero.
Up until now, things had been fairly straightforward. Then, once I hit the ‘two’ box, things began to get more complicated. Here, I couldn’t put four or three because two of the boxes had already been filled, and I couldn’t change that. Then, I tried two. This could work, but only if there was a two elsewhere. I couldn’t put a two in the ‘one’ box, but I could put it in the zero box, because of the ‘four’ and ‘three’ boxes. Good thing I didn’t change those. That left me with the ‘one’ box. There was really only one option for that box, and that was putting a one in it.
That was my process for solving the seemingly daunting, but surprisingly easy