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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.

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

1) 5 + 7 + 3 + 6 + 9 = 30, which is a multiple of 6.

2) 1 + 2 + 4 + 8 + 0 = 15, which is a multiple of 5.

3) 5 + 3 + 9 + 2 + 8 = 27, which is a multiple of 9.

4) 7 + 6 + 1 + 4 + 0 = 18, which is a multiple of 2.

5) 57 is a multiple of 3.

6) 36 is a multiple of 4.

7) 91 is a multiple of 7.

8) 24 is a multiple of 8.

9) 80 is a multiple of 10.

10) 19, 43, and 05 are prime numbers.

A) The sum of digits 0 through 9 is 45. To satisfy Conditions 1 and 2, a pair of multiples must equal 45. The only pair that satisfies those conditions are 30 (multiple of 6) and 15 (multiple of 5). Likewise, to satisfy Conditions 3 and 4, the pair of multiples must be 27 (multiple of 9) and 18 (multiple of 2).

B) Make lists of the multiples of 3, 4, 7, 8, and 10 (for Conditions 5 through 9). Delete multiples whose digits are the same (33, 44, 66, 77, 88, 99).

C) To satisfy Condition 9, J must be 0. Delete multiples that require 0 in a different location. For JA to be prime (Condition 10), A must be: 2, 3, 5, or 7. Delete multiples of 3 that have another digit for A.

D) For FE and HC to be prime (Condition 10), C and E can not be even or 5. Delete multiples of 4 and 7 that have another digit for C and E.

E) Per Condition 4, B + D + F + H + J = 18. D, H, and J must be even (required to form multiples of even numbers). To attain the sum of 18 (an even number), B and F must either be both odd or both even. If both are even, then the sum of those five digits would be 20. B and F must be odd. Delete multiples of 3 and 7 that have an even digit for B and F. The only remaining multiple of 7 is 91. E is 9 and F is 1. Delete multiples that require 1 or 9 in a different location.

F) The only remaining options for AB contain 7. Delete multiples that require 7 in a different location. The only remaining options for CD contain 3. Delete multiples that require 3 in a different location.

G) To satisfy Condition 1, CD must be 36 and AB must be either 57 or 75. C is 3 and D is 6. Delete multiples that require 5, 6, or 7 in a different location.

H) The only remaining options for GH contain 4. Delete multiples that require 4 in a different location.

I) Per Condition 3, A + C + E + G + I = 27. Since C =3 and I = 9, the other three must total 15. There is only one combination of remaining multiples (57, 24, and 80) that can satisfy this condition. A is 5, G is 2, and I is 8. B is 7 and H is 4.

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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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