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Category: | Logic |

Submitted By: | cnmne |

Fun: | (2.08) |

Difficulty: | (3.06) |

There is a ten digit mystery number (no leading 0), represented by ABCDEFGHIJ, where each numeral, 0 through 9, is used once. Given the following clues, what is the number?

1) If A > B, then C = 5 or 7, else C = 0 or 1.

2) If B > C, then D = 1 or 2, else D = 4 or 9.

3) If C > D, then E = 6 or 9, else E = 3 or 5.

4) If D > E, then F = 2 or 4, else F = 1 or 6.

5) If E > F, then G = 5 or 6, else G = 0 or 7.

6) If F > G, then H = 1 or 4, else H = 8 or 9.

7) If G > H, then I = 0 or 8, else I = 6 or 7.

8) If H > I, then J = 3 or 8, else J = 2 or 5.

9) If I > J, then A = 3 or 7, else A = 4 or 8.

10) If J > A, then B = 0 or 9, else B = 2 or 3.

Show Answer |

Posted by duchin38 | 11/28/11 |

I just did 0123456789, which is that number just not in that order....can that work? | |

Posted by oddrey | 04/19/12 |

It's been a while since I took Computer Science, but is it just me or is this teaser written in Java code? | |

Posted by dewtell | 01/27/15 |

Nice challenging teaser, with lots of possibilities to explore. Not sure if I follow your chain of possibilities to eliminate. I did it as follows: 1) Assume H=8, as you did. Leads to a contradiction, H can not be 8. 2) Assume H=9. This then implies FB, C>D, E=6, I=7, DF, G=5, and a contradiction, as that does not leave any legal values for C. So H can not be 9. 3) Therefore, F>G. G can not be 6 or 7. Assume that G=0 (the easiest way to get F>G). Then G F. E must be 9 to be > F. We must have C>D to have E be 9, so C must be 7, since it can not be 5. This implies A>B, and we know B | |

Posted by dewtell | 01/27/15 |

Lost a bunch of the analysis in the posting there. Anyhow, the next bit was showing that either B or D must be 9 under the G=0 hypothesis, and both of them lead to contradictions. So F=6 and G=5. Now I must be 0 or 8, 8 leads to contradictions, so I=0, and we get to the given solutions. |

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