*Spoiler for New Scientist Enigma 1722: “Double Latin” *(Follow the link to see the puzzle.)

Label the rows 1 through 5 (bottom to top) and the columns a through e (left to right) so that, for instance, the red side of card e4 is **3**.

Now examine red card sides to see what numbers they can have on them. At e1 and e5, for instance, we can have **1** or **2**. Let’s put **1** at e5 and **2** at e1.

Now d2 can only be **1** and d4 can only be **5**. From there d5 can only be 4, d1 must be **3**, c3 must be **3**. And so on; every card can have only one value and the complete red side is:

5 3 2 4 1 2 4 1 5 3 1 5 3 2 4 3 2 4 1 5 4 1 5 3 2

On the other hand if we put **1** at e1 and **2** at e5, we can again get a unique set of values for the red side:

4 1 3 5 2 5 2 4 1 3 1 3 5 2 4 2 4 1 3 5 3 5 2 4 1

Now consider the green side. In the first case, b1 has **2** on the green side and **1** on the red. None of the rest of row 1, column b, a2, or c2 can be **2** on the green side, but also neither can d2, a3, or c4 because they also have a **1** on the red side. This means e2 must be **2**, after which c3 must be **2**, and then there is nowhere in column d to put a 2. So that case is impossible.

In the other case, not only can none of the rest of row 1, column b, a2, or c2 can be **2** on the green side, but also neither can e2, c3, or a4 because they also have a **5** on the red side. So d2 must be **2**, and then a3 must be **2**. b4 must be **1**, and e3 must be **1**. Now e4 can’t be **2**, because there’s a **3** on the red side, and that combination has already been used at d2. So c4 and e5 are **2**. Likewise a1 and c2 are **1**.

That accounts for all the **1**s and **2**s. Now there’s a problem, because there’s nothing left to constrain the rest. So for example we can put **3** at a5, b3, d4, c1, and e2; **4** at a4, c5, b2, d3, and e1; and **5** at a2, b5, c3, d1, and e4. That satisfies the constraints. But we also could interchange all the **3**s with all the **4**s for another valid solution, or with all the **5**s for another — in fact there are six solutions corresponding to all the permutations of {**3**, **4**, **5**}.

So is this a problem? The enigma asks for “the numbers hidden under L A T I N”. I interpreted that as meaning the numbers on the hidden side — the green side. There is no unique answer if so.

But if this means the letters L A T I N are hiding the red-side numbers, and these are the numbers to be given, then there is a solution: 4 5 1 2 3. I suppose this must be what was meant, but if so I don’t like the wording of the question much.

Then again, I could be pedantic and say we *do* know the green side numbers; they’re 1 2 3 4 and 5. It didn’t say we had to specify them in any particular order!