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.

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

The 3-digit squares without duplicate digits are 169, 196, 256, 289, 324, 361, 529, 576, 625, 729, 784, 841 and 961. Of these, the two 3-digit squares are 256 and 625, and the three 3-digit squares are 169, 196, and 961. The common digit between the set of two and set of three is 6, therefore G is 6, (E, J) is (2, 5) or (5, 2), and (B, D) is (1,9) or (9,1).  The remaining somewhat unassigned digits at this point are 0, 3, 4, 7 and 8.

A 3-digit triangular number contains the values of A, C and F.

The 3-digit triangular numbers with no repeating digits are 105, 120, 136, 153, 190, 210, 231, 253, 276, 325, 351, 378, 406, 435, 465, 496, 528, 561, 630, 703, 741, 780, 820, 861, 903 and 946. As the A, C and F can not be 1, 2, 5, 6 or 9, this leaves 378, 703 and 780.

G,I and C, not necessarily in that order, are increasingly consecutive digits as are F, B, J and A.

If the 3-digit triangular number is 378, then H and I are, not necessarily respectively, 0 and 4. For G, I and C to be in consecutive order requires I=4, C=5, G=6 and H=0. Yet, 5 must be either E or J, therefore 378 is not the triangular number.
If the 3-digit triangle number is 780, then H and I are, not necessarily respectively, 4 and 8. For G, I and C to be in consecutive order, with 5 being E or J, requires G=6, C=7, I=8 and H=4. With F, B, J and A being consecutive digits, then F and A must be, not necessarily respectively, 7 and 8. As the digit I would need be 8, 780 can not be the triangular number.
This leaves 703. With H and I being, not necessarily respectively, 4 and 8. For G, I and C to be in consecutive order, with 5 being E or J, requires G=6, C=7, I=8 and H=4. As F, B, J and A must be consecutive, this requires the digits to be 0, 1, 2 and 3.
Therefore, F=0, B=1, E=2 and A=3..., and...

1. (A,B,C,D,E,F,G,H,I,J) = (3,1,7,9,5,0,6,4,8,2)

2. 1221 = 234 + 987 = (JAH + DIC)
           or 345 + 876 = (AHE + ICG)
           or 678 + 543 = (GCI + EHA)
           or 789 + 432 = (CID + HAJ)
Therefore the sets of 6 letters are (A,C,D,H,I,J) and (A,C,E,G,H,I). 

Posted by Dej Mar on 2009-12-20 05:22:22