The letters A to J have been assigned values for 0 to 9 but not in that order.

G,I and C, not necessarily in that order, are increasingly consecutive digits as are F, B, J and A.
A 3-digit triangular number contains the values of A, C and F.
In some order E, G and J form two 3-digit squares while B, D and G treated similarly would yield three 3-digit squares.

1. Determine the values assigned to each letter.
2. Determine which sets of 6 letters may be arranged so that they form two 3-digit numbers where the values in one rise consecutively, those of the second decrease consecutively and their sum is a 4-digit palindrome.

Part I
(E, G, J) form two 3-digit squares, and (B, D, G) form three 3-digit squares.

Crossing off those that contain duplicate digits:
100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961

From those remaining, look for those that don't have multiple values formed from the same three digits:

169, 196, 256, 289, 324, 361, 529, 576, 625, 729, 784, 841, 961

From these 5 bold values, we can see that (E, G, J) must be (2, 5, 6) in some order, and (B, D, G) must be (1, 6, 9). As G is in both sets, G = 6. Looking through the list of 3-digit triangular numbers, and crossing off those that contain multiple digits or any of the digits in (B, D, E, G, J):

105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990

So, (A, C, F) must equal: (3, 7, 8), (7, 0, 3), or (7, 8, 0). As A and F must be within the same 4-digit string (i.e. 0123, 1234, etc), and C must be within two digits of G, C must be either 4, 5, 7, or 8. That rules out 378 and 780 for our triangular number, leaving: (A, C, F) = (7, 0, 3), and C = 7.

While I'm not sure whether or not (F, B, J, A) are supposed to be in order, the values of A and F don't matter as they aren't used anywhere else that can allow us to determine for sure which one is which value. While they can be equal to (0, 3) in either order, just to be safe, we'll say A = 3, F = 0.

Since (E, G, J) is (2, 5, 6) and J must be within the string 0123 containing both A and F, then J = 2, and subsequently, E = 5. That eliminates A, F, and J from our four consecutive values (A, B, J, F), meaning B is the last remaining digit in that string, or B = 1. From our three squares containing the digits (1, 6, 9), we can see that D = 9. As (G, I, C) is either (5, 6, 7) or (6, 7, 8), and 5, 6, and 7 are already used, the last unknown value in that set must be I = 8.

We now know that: A = 3, B = 1, C = 7, D = 9, E = 5, F = 0, G = 6, I = 8, and J = 2, leaving our only unknown to be H = 4.

Part II
As the set of 6 numbers must be able to be split up into two 3-digit values, one decreasingly consecutive, and one increasingly consecutive, we know they can't overlap. Also, in order to be a palindrome, the first and last digits of the sum must be 1, and as 0 is usually considered to be low, that leaves us with the last digits being:

(9, 2), or (8, 3) -- with corresponding sets of 6 being (7, 8, 9, 4, 3, 2) and (6, 7, 8, 5, 4, 3). Both of which sum to 1221. So, assuming 0 is only low, there are two sets of values that can be divided into two 3-digit numbers that fit the properties given.

If 0 could also be used as a high number, we would have three sets of values that would work, rather than two.

Edited on December 19, 2009, 5:52 pm

Posted by Justin on 2009-12-19 16:58:33