## 3.2: Numbers Upon Numbers (Solution) |
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First, you need to realize that each of the 25 three-digit numbers fits into a different box based on its digits, and therefore each number has to have a certain format.

The logic for finding the twenty-five integers is given below. Once they are found, the numbers need to be placed in the grid. Many teams had written a large five-by-five grid of letters and stared at it, confused. We told them to go back and look at the web page, which was often enough of a hint. The three-digit numbers themselves, not their identifying letters, needed to go into the grid. The digits that mark the rows should be written in black and the digits marking columns should be written in that brownish color, as was done for the number W. When that's done, you'll see a bitmap that looks like this:

_X_ _XX X_X XX_ XX_ X_X X__ X_X X__ X_X X_X X__ XXX XX_ XX_ X_X X__ X_X X__ X_X _X_ _XX X_X XX_ X_XThe answer is OCHER, explaining the choice of colors for the bitmap.

And now, on to finding out the numbers for the grid...

There are far, far too many answers if the numbers are unconstrained. The fact that each number must belong in its own cell helps things along tremendously. Many solvers started by thinking that each three-digit number had to have two different digits, one from {1,2,3,4,5} and the other from {6,7,8,9,0}. (The example, 656, fits this description.) This is not correct, since it's possible that one number could have the same digit repeated three times. For example, if the final solution included 991, 992, 994, 995, and 999, then the position of 999 in the grid would not be ambiguous. (There would be no reason to say that "although it might not be exactly clear in the middle" where a number could go unless this were a possibility; otherwise, everyone would immediately know where every number went.) Fortunately, a solver can get fairly far thinking that multiples of 111 are illegal before running into one.

The best place to start is with A, I, and Q. We know that, at most, only five numbers can start with 0, and therefore Q > I > 100. Looking over a list of square numbers, we see that this only leaves us with two options for Q, namely 400 and 484. This means that 100 < I < 400, so 033 < A < 133. Trying out the possible values for A (040, 044, 050, 055, 100, 101, 110, 111, 116, 117, 118, and 119), we see that the only values for A and I that work are A = 100, A = 101, A = 110 and A = 111. Now we know that I starts with 3. At most five numbers can start with 3, so we can get a lower bound of 400 <= N. Since 400 <= N < Q, we can fix Q at 484.

Continuing with N, we now know that 400 <= N < 484. This translates into 200 <= D < 242. The largest well-formed value for D in that range that is well-formed is 229, so we can say D <= 229. Popping back to N, we get N <= 458, which we can similarly tighten to N <= 449. We can even slide that down to 448 since we're about to divide by two again. We find that D <= 224, which we tighten to D <= 222. At this point, we can't limit our range any more. We are left with four options for D (200, 202, 220, and 222), but we can use the rule for G to cut that down even more. Trying out Q - D = G on the four options and looking for decent values of G, we can cut our set of options in half: D = 202 or D = 222.

We now know that 100 < C < 202, and since C is also prime, there are only three numbers that fit the bill: 181, 191, and 199. We can try out these three combinations for C against the two combinations for D (and G and N) to get six stabs at the rule U = C + N. The result is that C is either 181 or 191 and that D must equal 202. With D found, we also can find G = 282 and N = 404.

We now look at F, a prime between 202 and 282. There are only three primes in that range (227, 229, and 277), and only one gives us a good value for T when we use the rule T = F + G. Therefore, F = 277 and T = 559. Since T fills the 5/9 slot, we have knocked out one of two options for U. It follows that U = 585 and C = 181.

Now turn your attention to E, snugly placed between D and F. From the rules for O and J, we know that E must be even; we also know that E must go in 2/6 or 2/9. This leaves us with 222, 226, or 262. Only 262 gives us a well-formed O, so we have E = 262 and O = 464.

Take a deep breath. We could have reached this point without realizing that multiples of 111 are legal. We would have one fewer option for A and I, but everything else would have worked just fine. (This is left as an exercise for the reader. No, really, it's true, straightforward to show, but too much of a sidetrack to bother with here.) Now we are ready to calculate J = (D + O) / 2, and we see that J = 333. Solvers at this point many panic, but there really is nothing wrong here. Computer programs will just reject this option and move on, finding no other possible solution. Muahaha.

Continuing, looking at U and W and a list of vacant squares, we know that V must be 611, 616, 633, or 636. Using the rule V = F - K, we see that V = 616 and K = 339. Using K = (B + T) / 2, we find that B = 119.

Because M = (F + P) / 2, we know that P must be odd. Knowing O, Q, and a list of vacant squares, we can deduce that P = 477. It follows that M = 377.

With D = (A + B + M) / 3, we can finally pin down A = 110 and I = 330. Next we find that X = B + E + I = 711, that H = (C + K + M) / 3 = 299, that L = (A + V) / 2 = 363, that S = (H + X) / 2 = 505, and that Y = A + G + L = 755.

We're left with R, a prime number that must go in the 4/9 slot; it must be 499.

To summarize,

A: One-third of I = 110 B: Just some number = 119 C: A prime number = 181 D: The average of A, B, and M = 202 E: The sum of A and X, less T = 262 F: Another prime number = 277 G: The difference of Q and D = 282 H: The average of C, K, and M = 299 I: Thrice A = 330 J: The average of D and O = 333 K: The average of B and T = 339 L: The average of A and V = 363 M: The average of F and P = 377 N: Twice D = 404 O: The sum of D and E = 464 P: Another pesky number = 477 Q: A square number = 484 R: Yet another prime number = 499 S: The average of H and X = 505 T: The sum of F and G = 559 U: The sum of C and N = 585 V: The sum of F and K = 616 W: Six hundred fifty-six = 656 X: The sum of B, E, and I = 711 Y: The sum of A, G, and L = 755