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Indexed Puzzle (Posted on 2004-07-19) Difficulty: 4 of 5
Here is a numbered list of statements, some true, some false, which refer to a specific number (unique positive integer, base 10).

It just so happens that if a statement is true then its index number appears among the number's digits, and if a statement is false then its index number does not appear among the number's digits.

  1. The sum of the number's digits is a prime.
  2. The product of the number's digits is odd.
  3. Each of the number's digits is less than the next digit (if there is one).
  4. No two of the number's digits are equal.
  5. None of the number's digits is greater than 4.
  6. The number has fewer than 6 digits.
  7. The product of the number's digits is not divisible by 6.
  8. The number is even.
  9. No two of the number's digits differ by 1.
  10. At least one of the number's digits is equal to the sum of two other digits. (Any of the digits may be equal, as long as all 3 digits are distinct... for example: {2, 2, 4} or {2, 3, 5} )
Find the number.

See The Solution Submitted by SilverKnight    
Rating: 4.3750 (8 votes)

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Solution Solution | Comment 5 of 22 |
If 1 is a digit, the product of all digits is odd, so all digits are odd, so the difference between any two digits is 2 or more, so statement (8) is true, but then 8 would be a digit... 1 isn't a digit, and there is at least one even digit. Possible digits so far: 0, 2, 3, 4, 5, 6, 7, 8, 9.


If 6 is a digit, the product of the digits is a multiple of 6, so statement 6 is false... 6 isn't a digit. Possible digits so far: 0, 2, 3, 4, 5, 7, 8, 9.


If 4 is a digit, since 1 isn't a digit, by statement 4 only 0, 2, 3, and 4 can be digits, so 7 cannot be a digit, so by statement 7 the number must end in 3. Since statement (3) would be true, the number would have at most four digits, so statement (5) would be true, and there would be a 5 in the solution... 4 isn't a digit. Possible digits so far: 0, 2, 3, 5, 7, 8, 9.


If 7 is a digit, 8 cannot be a digit since then statement (8) would be false. If 8 isn't a digit, there must be two digits differing by 1, and the only possibility is 2 and 3, so those two statements are true. From statement (2), 0 cannot be a digit, and from statement (3), all digits are different, so only 2, 3, 5, 7, and 9 would be left, and the number must start with 2, and end with an odd number, so statement (7) would be false... 7 isn't a digit. Possible digits so far: 0, 2, 3, 5, 8, 9.


If 5 isn't a digit, the number has six or more digits. If 3 is a digit, all must be different, but we have only five possibilities, so 3 isn't a digit either, and we are left with only 0, 2, 8, and 9. 2 cannot be a digit, for in that case 3 would also be a digit, so we are left with 0, 8 and 9. Since the number is odd, 9 is in, and then 8 is out, and only 0 and 9 are left, but they don't differ by 1... 5 is a digit. Possible digits so far: 0, 2, 3, 8, 9. Sure digits: 5.


If 8 isn't a digit, 2 and 3 must be digits, and statement (2) implies 0 cannot be a digit, so we are left with 2, 3, and 5 (all for sure), and, since 2+3=5, also 9 for statement (9) is true. As statement (2) is true, the number would be 2359, but that makes statement (1) true... 8 must be a digit. Possible digits so far: 0, 2, 3, 9. Sure digits: 5, 8.


As 8 is a digit, 9 (and 7) cannot be digits, because of statement (8). Possible digits so far: 0, 2, 3. Sure digits: 5, 8.


If 2 is a digit, then the number ends in 8 because of statement (2), but then the number would be even, so 7 would be a digit... 2 isn't a digit. Possible digits so far: 0, 3. Sure digits: 5, 8.


If 3 is a digit, since 3+5=8, 9 would be a digit, but it isn't... 3 isn't a digit. Possible digits so far: 0. Sure digits: 5, 8.


For the product of numbers to be divisible by 6, 0 must be included, since 3, 6, and 9 have already been excluded. Sure digits: 0, 5, 8.


Since there is no 3, statement (3) is false, and there must be some equal digits. There cannot be two 5's or two 8's, for that would make statement (9) true, so there must be two 0's, and one 5 and 8. There cannot be three 0's either, for that would also make statement (9) true.


The number is odd as we saw, so it must end with 5, and if it doesn't start with 0, it must be 8005.
  Posted by Federico Kereki on 2004-07-19 16:14:07

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