Each of the capital letters in bold denotes a different base ten digit from 0 to 9 to satisfy each of the following alphametic relationships. None of the numbers can contain any leading zero.
(i) ONE is a perfect square and:
(ii) Precisely one of TWO, THREE and FOUR is a perfect square, and:
(iii) Precisely one of TWO + 1, THREE + 1 and FOUR + 1 is a perfect square, and:
(iv) Precisely one of TWO + 2, THREE + 2 and FOUR + 2 is a perfect square.
Determine the number that is represented by FORTUNE.
Method:
I did this in Excel and wrote out all combinations of sqares, squares - 1, and squares - 2 that make three, four and five digit numbers and I divided these out into separate worksheets. I also made a chart of the digits 0-9 and FORTUNEWHZ so I could look at the eliminations easier (ala standard Logic Puzzles).
(1) ONE: Eliminate all combinations of letters that cannot correspond to numbers. (e.g. in this case have any numbers that repeat regardless of digit location)
(2) "O" cannot be 0 or 4.
(3) "N" cannot be 0, 1, or 3.
(4) TWO: Eliminate all combinations of letters that cannot correspond to numbers.
(5) "T" cannot be 0.
(6) "W" cannot be 0 or 1.
(7) Eliminate all instances of "0" that cannot be possible (0, 4).
(8) THREE: Eliminate all combinations of letters that cannot correspond to numbers.
(9) Eliminate all instances of "E" that cannot be possible (0, 2, 3, 7, 8).
(10) "T" cannot be 2.
(11) TWO: Eliminate all instances of "T" that cannot be possible (2).
(12) THREE: "H" cannot be 1 or 5.
(13) FOUR: Eliminate all combinations of letters that cannot correspond to numbers.
(14) Eliminate all instances of "O" that cannot be possible (0, 4).
(15) "F" cannot be 0.
(16) At this point I consolidated the slimmed down list of all four numbers into one spreadsheet to compare values and also to see if they are perfect squares, +1 squares and +2 squares.
(17) I noticed that THREE had only values ending in 4 for all of the perfect squares it produced. I then made an assumption that E was 4 so that even if it wasn't, I could then eliminate THREE from being a perfect square. After starting the snowball with E being 4, I came to the correct answer. The logic chart really made things easy for this.
(NOTE: I stopped after arriving at a correct answer, but there could be other possibilities--although with such few numbers left I sincerely doubt it.)
Solution:
ONE: 784 (28^2) (i)
TWO: 167 (+2 = 13^2) (iv)
THREE: 19044 (138^2) (ii)
FOUR: 3720 (+1 = 61^2 + 1) (iii)
and therefore:
FORTUNE: 3701284
Edited on November 19, 2009, 3:39 pm
Logic and a little trial & error:
