# Math Puzzle for Quarantining Humans: 4444

This post is part of a series of math puzzles. Here is the list of puzzles for reference:

Now, let’s continue with 4444:

# 4444: PROBLEM

You may have seen this kind of problem before in 3333. Now, 4444 is kind of an upgrade in difficulty compared to 3333. If you didn’t see it yet, I’ll just describe the problem below (although I’d attempt 3333 before attempting this one because it is much harder!).

The problem goes as follows: you have one number minus another number, and the result is 4444. The question is: what are the 2 numbers that give this result? But here’s the catch: the digits of the two numbers must be chosen from 1–9, and each digit must be used at exactly once.

For instance, `5678 — 1234` does not work because it does not use the digit `9`. `12345 — 7899` is invalid because the result is not `4444`, it uses 9 twice, and it doesn’t use 6.

Similar to 3333, I already gave you a hint above that the first number must have 5 digits and the second one has 4 digits. That’s the only way this could work. Try it!

Scroll down to see a hint if you get stuck, and scroll further down to see the solution.

# 4444: HINT #1

Have you attempted 3333? That should give you a bit of a hint as to where to start.

Scroll below for another hint…

# 4444: HINT #2

Hmm… Maybe attempting the other two prior problems in this series would help. Have you tried both of them?

Scroll below to see the solution when you’re ready!

# 4444: SOLUTION

There’s a reason why this problem was posted after Slicing the Parabola, and that reason is, there are no solutions to this problem!

However, if you attempted Slicing the Parabola, you’ll know that stating that there are no solutions is not enough because we must then prove that there are no solutions. In Slicing the Parabola, we showed that there are no solution through a logical contradiction. In 4444, we’ll follow a different approach: that of deductive reasoning.

But before that, let’s try to see what you might have attempted when solving this. Similar to 3333, let’s list all digit pairs that result in `4` :

• `5-1`
• `6-2`
• `7-3`
• `8-4`
• `9-5`

Again, we can’t use the very first one because `1` must be the first digit of the first number (this follows the same logic as to why the first digit in the first number of 3333 must be a `1` ).

Then again, we must also think about carrying. If we were to carry only once, we would need to have some digit, plus 10, and minus some other digit, the result of which is `4` . In that case, we have:

• `10+1-7` , giving us the`1-7` pair, but we can’t use this because `1` is already used
• `10+2-8` , giving us the `2-8` pair. Later, I’m going to write this as `(10)2-8`
• `10+3-9` , giving us the `3-9` pair. Later, I’m going to write this as `(10)3-9`

Now, if like in 3333 you tried the pairs that we have (`2–8` , `3-9` , `6-2` , `7-3` , `8-4` , and `9-5` ), you’ll quickly see that it won’t be as successful as before. If you take `2-8` , then you can’t take `3-9` (because there’s no more carry opportunities of the sort), you can’t take `6–2`, and you can’t take `8-4` (because 2 and 8 are used in the `2–8` pair), leaving you with only 3 choices. Similarly, if you take `3-9` , then you can’t take `2-8` , `7-3` , and `9-5` , leaving you with only 3 choices again.

It seems we’re stuck here. However, we could go one step further and do a double carry. That is, we carry the one (first digit of first number) and then from the carried `10` , carry once again (leaving that first carry to be a `9` and getting another `10` ). This means that now, in addition to the “10 + digit — digit”, we also now have “9 + digit — digit”:

• `9+1-6` , giving us the `1-6` pair, but we can’t use this because `1` is already used.
• `9+2–7` , giving us the `2-7` pair. I’m going to write this as `(9)2-7`
• `9+3–8` , giving us the `3-8` pair. I’m going to write this as `(9)3-8`
• `9+4–9` , giving us the `4-9` pair. I’m going to write this as `(9)4-9`

Note that if we do a double carry, we must also use one of the `10` and something pairs. Here, we get many possibilities, but only the following pairs work for a double carry:

• `(9)2-7` and `(10)3-9`
• `(9)4-9` and `(10)2-8`

We can’t use `(9)3-8` anywhere because it uses up all of our `(10)` pairs digits.

However, you can notice that even now, we still are stuck. Note:

• If we use `(9)2-7` and `(10)3-9` , then the only original pairs left is `8-4` .
• If we use `(9)4-9` and `(10)2-8` , then the only original pairs left is `7-3` .

It seems we’re out of luck again. Another strategy here would be to try triple carry or even quadruple. However, one thing you note is that even if you carry more than twice, the number you can add would still be `9` at the very least. You don’t drop down to `8` . Meaning that the maximum pairs that we’d be left with are the above, which we just proved that they leave us with no working possibilities.

From here, we may say that there are no solutions…

## But, is that sufficient of a proof that there are no solutions?

I don’t think so. We showed that there are no solutions if we follow this carrying strategy. However, could it be possible that our strategy was just wrong? That there could be some other magical strategies that work?

Because we don’t know that, what if we instead tried to show that there are no solutions to this problem?

## Proof that there are no solutions

There probably are many ways to do this. However, here’s how I approached it: I used a set of logical conclusions that lead to a much easier problem to solve that proved to be impossible, and that led to this problem being impossible.

We start from the conclusion that the digit `1` is the first digit of the top number. See 3333 for how we deduced that.

Then, let’s think about the digit `5`. First, we notice that `9–5` (which has `5` at the bottom) represent a pair of digits that lead to `4` . Which lead us to the following conclusion:

If 9 is on top, 5 must be at the bottom

We know that condition to be true because there is no way to carry when `9` is on top. That is, if we carry from `9`, our problem would lead to carrying twice, meaning that we’d use all two of our 10 pairs. One of those pairs needs `9` at the bottom, which would contradict that `9` was on top to begin with. Which means that the digit at the bottom must be something that would lead to `4`, and only `5` works in that case.

Another conclusion we have is:

If 5 is at the bottom, 9 is on top

This is sort of saying the same thing as above except from a different angle. One statement doesn’t imply the other so we must prove both. This one is to say that if we have `5` at the bottom, there is no other way to get `4` unless we put `9` on top. That is because we already showed that by carrying, which stops at adding `9+something-something` , we can’t have `5` at the bottom. So, the only possible way left to get a result of `4` is by having `9` on top.

Next, what if `5` was on top? Similarly, we can’t carry with `5` on top because that would lead to at least `5` as a result. That is, the result of `9+5-9=5` . `5` is as small as it gets if we carry and have `5` on top. This means:

If 5 is on top, then 1 is at the bottom

However, we can’t use this because `1` is already used. This implies `5` cannot be on top because it has to be paired with `1` which is already used.

So, the above implies that the pair `9-5` must be in the solution. Also notice that `9-5` is an isolated pair (i.e., it can be placed anywhere-ish ). So, for that reason, let’s just place it at the end.

The next deduction that we’re going to make is:

We have to carry at least once

This is because in order to get rid of that `1` in the top number, we have to carry. This implies that we’re going to have at least one`(10)x-y` pair. If you go back, you’ll notice that we have 3 of those: `(10)1-7` , `(10)2-8` , `(10)3-9` . However, we can’t use `(10)1-7` because `1` is already used, and we can’t use `(10)3-9` because `9` is already used (by the `9–5` pair above). That leaves us with the `(10)2-8` pair, and this implies that the the pair `2-8` is going to be used. Also, note that we can carry in this way only once because after using `2-8` , there will be no more `(10)x-y` pairs.

The next question is, where exactly is this `2-8` pair? It has to be somewhere after the `1` . Here, we conclude that it’s literally right after the `1` . The reason is that if it wasn’t there, we’d have to do a double carry with a `(9)x-y` pair. However, we can’t do that because if you go back to our `(9)x-y` pairs above, they are: `(9)1–6`, `(9)2-7` , `(9)3-8` , and `(9)4-9` . They all can’t be used because all their digits are already used (namely, `1`, `2`, `8`, and `9`).

This makes it such that our original problem is reduced to the following problem:

Two two-digit numbers, one minus the other, such that the result is `44` . Except, we can only use the digits `3`, `4`, `6`, and `7`. From here, we can make some simple conclusions. There is no way to carry and still get `4` . That is because none of the available digits result in a carry pair (as we showed with `(10)x-y` pairs and `(9)x-y` pairs). Therefore, we can only use the present digits. Amongst them, the pair `7-3` gives us `4`:

That means we’re left with the following even simpler problem:

One digit minus another, and the result is `4` . However, we can only use two digits: `4` and `6` . You can now clearly see that this simpler problem is impossible to solve, which means that the general problem of using the nine digits is also impossible!

That finally shows that there are no solutions to this problem!

# 4444: MORAL

This is yet another lesson about problems that show no solution. The key takeaway here is that to show that there are no solution, you must tackle the problem of showing that there are no solution to the problem instead of showing that all possible strategies lead to no answers. That is because showing that all possible strategies lead to no answers implies that you’ve tried all possible strategies, and unless you prove that you have tried all possible strategies (which is VERY difficult in math as there is always a millions way to solve a problem), you can’t use that type of no-solutions-proof.

Of course, the other smaller lesson is that if all your strategies don’t work, maybe it’s time to ask whether a solution exists.

With that said, I’ll see you on the next one!

# End Notes

 I have to say “anywhere-ish” because that’s not necessarily true. In case we do carry overs, it’d mean that some digits depend on others (like they did in 3333). Therefore, this is not actually true, but at the same time, it’s why it makes it safe to put this pair at the end.

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