Math Puzzle for Quarantining Humans: 3333
My next few posts will be a series of simple math puzzles that will require a little bit of cleverness to solve. Here’s the list if you want to jump to a particular problem:
- 3333 (this post)
- 6 Inscribed Circles
- Slicing the Parabola
- 4444
- […coming soon…]
- […coming soon…]
Hopefully, some bored human in quarantine might find joy in solving these problems. Let’s start with 3333:
3333: PROBLEM
You have one number minus another number, and the result is 3333. 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 — 2345 does not work because it does not use the digit 1 and it uses the digit 5 twice. 12345 — 6789 is invalid because the result is not 3333 .
Now, I already gave you a hint in the picture above that the first number must have 5 digits and the second number 4 digits. That’s the only way this would work. Now, you have to find the two numbers. GO!
Scroll down to see a hint if you get stuck, and scroll further down to see the solution.
3333: HINT
There’s a high possibility that you figured this out already, but here it goes. Since the operation’s result (one number minus another) is 3333, a positive number, the top number must be larger than the bottom number, and since we need to use all 9 digits, the first digit of the top number must be a small one (because if it’s too large then the result will be too large). In fact, like mentioned above, it’s 5 digits and the bottom is 4 digits. Thus, the first digit must be a 1. It can’t be 2 because 20,000–9,999 = 10,001 (the smallest difference between a 5 digit number starting with 2 and a 4 digit number).
3333: SOLUTION
It turns out that there are many solutions to this problem, and here they are:
12678 - 9345 = 3333
12687 - 9354 = 3333
12768 - 9435 = 3333
12786 - 9453 = 3333
12867 - 9534 = 3333
12876 - 9543 = 3333If you look closely, the solutions resemble each other a lot, and they reveal something about how you find the solution.
First, we know we’re doing something of the form 1xxxx — xxxx = 33333 . Because we know that 1 is the first digit (as explained in the hint), we know that we must carry in order for that 1 to disappear. Let’s keep that information aside for now.
We need to list all the possible ways we get get 3 from subtracting two digits. We get it with 4-1 , 5-2 , 6-3 , 7-4 , 8-5 , 9-6 . We can’t use 4-1 because the digit 1 is already used. At the same times, we can use at most 3 of these because after 3 selections of pairs of digits, other digits start repeating (this is because any group of more than 3 pairs will repeat a digit).
It seems like we’re out of luck. However, remember that we have to carry over to get rid of the 1 . Which means, there are other possibilities at play. If we carry, we can get 10 plus whatever digit was there before. All the ways we can get 3 from 10 plus a digit minus another digit are: 10+1-8 and 10+2-9 . Those give us 1-8 and 2-9 pairs. However, 1 again cannot be used, so we’re left with only one pair: 2–9. However, that’s just enough to get one of the solutions. We can use the following pairs: 2-9 , 8–5 , 7-4 , 6-3 . With that, we get 12876 — 9543 = 3333.
Now, you can see that the order of the last 3 pairs that I gave has nothing inherently special. That is, the last 3 pairs could be permuted. The permutation formula is the following:
What this formula essentially describes is the following. Let’s say we have n objects, and let’s say we could order these objects into r possible placements. Then the total number of orderings is given by P(n,r) . Here’s a video that probably explains it better. With that said, P(3,3) = 3!/(3-3)! = 3!/0! = 6 . If you look above, we have exactly 6 solutions! So, that checks out with the solutions we gave above :D.
If you were able to solve this, why don’t you try 4444? It’s the same problem, except the result is 4444. It’s much harder than this one, so get ready!
