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Category: | Logic |

Submitted By: | cnmne |

Fun: | (2) |

Difficulty: | (3.29) |

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) A + B + C + D + E is a multiple of 6.

2) F + G + H + I + J is a multiple of 5.

3) A + C + E + G + I is a multiple of 9.

4) B + D + F + H + J is a multiple of 2.

5) AB is a multiple of 3.

6) CD is a multiple of 4.

7) EF is a multiple of 7.

8) GH is a multiple of 8.

9) IJ is a multiple of 10.

10) FE, HC, and JA are all prime numbers.

Show Answer |

Posted by wearymachine | 08/10/10 |

hah...this one took me a whole evening to solve...was up till 2 last night and now it's 7:50 AM and i have to go to work =(... was it worth the time??...YESS...amazing puzzle... | |

Posted by Stormer | 05/15/14 |

I think I found a possible alternative solution: http://imgur.com/aYmFxJg | |

Posted by sarggames | 10/28/14 |

i solved in less time and a very neat solution :lol: :D | |

Posted by sarggames | 10/28/14 |

nice teaser | |

Posted by javaguru | 11/10/15 |

I went about solving it differently. 1) J = 0 because IJ is a multiple of 10. 2) D & H are even (because CD and GH are even) and E and C are odd and not 5 because FE and HC are prime, and neither F nor H is 0. 3) C+D+E is even, so A+B is even since A+B+C+D+E is even. D+H+J is even, so B+F is also even since B+D+F+H+J is even. This means A, B & F must all be even or all be odd. Since three even numbers are already accounted for, A, B & F are odd, making G and I even. 4) E and F are odd and EF is a multiple of 7, so there are two possibilities for EF: 35, 91. G and H are even and GH is a multiple of 8, so there are three possibilities for GH: 24, 48, 64. F+G+H+I+J is a multiple of 5. F+G+H+J is one of 5+2+4+0 = 11, 1+2+4+0 = 7, 5+4+8+0 = 17, 1+4+8+0 = 13, 5+6+4+0 = 15, or 1+6+4+0 = 11. I is even, so to get a multiple of 5 by adding I to each of those, I has to be 4, 8, 8, 2, 0, or 4, respectively. Only two combinations for FGHIJ are possible: 12480 and 14820. Both these have F = 1, so F = 1 and E = 9. Also, since GHI = 248 or 482, D = 6 (the only remaining even number). 5) A is 3, 5 or 7 since A is odd and 0A is prime. A+B is divisible by 3 and B is odd and not 1 or 9. The possibilities for AB are 57 and 75. This means C = 3, the only remaining odd number. 6) A+C+E+G+I is a multiple of 9. C+E = 12, so A+G+I = 15 (because A is at least 5, A+G+I can't be 6). The possibilities for A+G+I are 5+2+8 = 15, 7+2+8 = 17, 5+4+2 = 11, or 7+4+2 = 13. So A = 5, G = 2 and I = 8. This leaves B = 7 and H = 4. Took less than ten minutes. 8) | |

Posted by saska | 03/06/17 |

Thanks for your puzzle. I managed to solve it with glpsol (a free linear and mixed integer programming solver, available for most platforms). If anyone is interested, I have uploaded the model file to pastebin: http://pastebin.com/8Dv1MWSe Have fun! |

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