 Strap on your thinking caps, because there are 20 blanks to fill in, and only one way to do it right.

Growing up as a youngest child, I often found it sort of frustrating that everyone else in the family knew more about the world than I did. That said, I don’t think I’d have felt any smarter if I’d had a younger brother like Japanese Twitter user Aty’s.

Aty (@aty_carbuncle) recently shared a math puzzle created by his younger brother, who’s in the first year of junior high school, meaning he’s 12 or 13. Little Bro wanted Big Bro to fill in the blanks on a simple long division problem…or at least it would have been simple if it didn’t have 20 blanks to fill in! Yep, aside from the 6 and the 0 in the computation section, and the decimal point in the quotient, everything else is up to you. With so many blanks, you might think there’re dozens of possible answers, but it turns out there’s only one possible solution.

I’ll give you a few seconds to try to figure it out/curse loudly in frustration at your screen.

All good? Okay. Like I said, there’s only one way to answer this question, and Aty, who was up to the challenge, explains it here. Of course, this is a math problem, not a linguistic one, so let’s walk through Aty’s steps in English. Before we get started, though, let’s rewrite the problem using Latin letters for each unknown digit instead of the Japanese katakana text from Aty’s handwritten notes. Because of the problem’s complexity, for notation purposes we’re going to treat each variable simply as its digit. For example, AB is a two-digit number, not the product of A times B. Also, it’s possible that some of the unknowns will end up being the same number (A and B could both be 1, for example).

Step 1: You actually need to start at the end, since the given zero is the key to the whole thing. Since T is being added as the quotient moves past the decimal point, T has to be zero, and in order to be the end of the calculation, so does V. Knowing that, let’s update the puzzle. Step 2: Going back to the original setup, AB x E = UV, and since we figured out that V = 0, we can also say that AB x E = U0, and that in turn means that the ones digit of B x E has to 0. If B x E gives us a 0 for the ones digit, the possibilities for B and E are (2, 5), (4, 5), (6, 5), (8, 5), (5,2), (5, 4), (5, 6), and (5, 8).

B can’t be 5, though. Because of the 6 we’re given at the start of the problem, we know that AB x C = KL6. If B were 5, then AB times anything could only end in a 5 or 0. Since B can’t be 5, we can cross off half the combinations, and now the possibilities for B and E are (2, 5), (4, 5), (6, 5) and (8, 5).

But wait, we know that AB x E = U0, right? If B isn’t 5, then the only way AB x E = U0 is possible is for E to be 0 or 5. But E can’t be 0, because it’s being added after the decimal point to complete the division. You wouldn’t need to do that if E were 0, so E has to be 5 (as for B, we’re going to come back to it later). Step 3: Now let’s figure out A. What would happen if A were 2 or larger? Since E is 5, A being 2 or more would mean that AB x E, or AB x 5, would end up as at least 100, if not more. In any case, it’d be a three-digit number. But we know that AB x 5 = U0, a two-digit number. So A can’t be 2 or larger, and it also can’t be 0, since it’s the largest digit in the divisor (i.e. there’d be no point to writing AB to start the question if A was 0). So by process of elimination, A has to be 1. Step 4: Next up, we know MNP – QR = S (remember, the 0 next to S, which started out as the unknown T, was only added as a drop-down when the calculation got past the decimal point). So we’ve got a three-digit number (MNP), and when we subtract a two-digit number (QR), the difference is a one-digit number (S). This eliminates a lot of possibilities.

QR is a two-digit number, so it can’t be any bigger than 99, which means MNP can’t be any bigger than 108. If it were, subtracting 99 wouldn’t give us a one-digit number. At the same time, MNP obviously can’t be any smaller than 100. So while P is still a mystery (it has to be something between 0 and 8, though) right now, we can be absolutely certain that M is 1 and N is 0. Step 5: Likewise, we can figure out that QR has to be at least 91, because if it were any smaller, subtracting it from a three-digit number wouldn’t give us a one-digit number. Once again, R is a momentary mystery, but at least we know that Q is 9, and we know that no matter what R eventually ends up being, it can’t be 0 (since we need to subtract at least 91 from the smallest possible three-digit number, 100, to get a one-digit one). We’re now about half-way through the solution, so let’s take a short break for a nice cup of tea. Okay, let’s finish this thing off! Remember, here’s what we’ve figured out so far. We also know that P is some number between 0 and 8 (i.e. it can’t be 9), and R is some number between 1 and 9 (it can’t be 0).

Step 6: Hey, remember all the way back in Step 2, where we found that the possibilities for B and E were (2, 5), (4, 5), (6, 5), (8, 5), (5,2), (5, 4), (5, 6), and (5, 8)? We eventually figured out that E has to be 5, so the remaining possible B and E combinations are (2, 5), (4, 5), (6, 5) and (8, 5). In other words, B has to be 2, 4, 6, or 8, so which one is it?

We know 1B x C = KL6, 1B x D = 9R, and that 9 has to be greater than 0. So let’s plug in the possibilities for B and see what happens.

● If B is 2, then we’d have 12 x C = KL6. But the only way 12 times a single-digit number would give us a three-digit number would be 12 x 9, which is 108. That doesn’t fit with KL6, so 2 is out as a possibility for B.
● If B is 6, we’d have 16 x C = KL6. Once again, there’s no value for C that works (the 16 x C three-digit possible values are 112, 128, and 144; once again, none have a 6 in the ones digit). So we can cross off 6.
● If B is 8, initially things look pretty good. 18 x C = KL6 works if C is 7 (the product becomes 126). But what about 18 x D = 9R? There’s only one way to multiply 18 by a single number and get a product with a 9 for the tens digit, and that’s to multiply it by 5, which gives you 90. But remember, back in Step 5 we figured out that R can’t be 0, so we have to eliminate 8 too.

▼ Anybody need another cup of tea? That makes 4 our last remaining hope for B. Fingers crossed, let’s see if it meets all the requirements (1B x C = KL6, 1B x D = 9R, R has to be greater than 0).

Sure enough, 14 x C = KL6 works if C = 9 (the product is 126), and 14 x D = 9R works if D = 7, which gives us the product of 98, in which case R is 8, clearing the “R can’t be 0” hurdle! What’s more, those are the only possible combinations for B, C, D, and R, so now we can plug all four of those numbers into the puzzle together. At this point, we’ve still got nine blanks to fill in, but now that we’ve got the complete quotient and divisor, the rest is a snap. 14 x 97.5 has to equal FGHJ, so those digits can only be 1365. If you’ve come this far, I’m guessing you don’t need any refreshers on how to do the basic division that’s left, so let’s turn things back over to Aty for his handwritten final answer. Basically, solving the puzzle requires you to be a mathematical Sherlock Holmes, eliminating the numerical impossibilities one by one until you’re left with the one and only truth. “This was a well-designed question. Nicely done,” Aty later tweeted, showing he’s as skilled at understatement as he is puzzle-solving. And as mentally satisfying as it was to tag along on the path to the solution, it’s also scary to think of what sort of brain teasers the younger math whiz will come up with next.