DFM

Solution to problem number 1.

Thu Sep 14, 2017 1:57am

2602:30a:2c0c:99d0:8829:2481:77d7:5c

Each age must be a divisor of 72 and here they are:

1,2,3,4,6,8,9,12,18,24,36,72

Call the ages of the three kids A, B, & C and, without loss of generality, assume that A <= B <= C (It is always possible to arrange it thusly.)

Table below shows all possible combinations.

Since knowing the house number is not enough, clearly there are two combinations that have the same total A+B+C, which are the two lines ending in 14, that is, lines 10 and 11. Since there is an oldest child, the top two kids cannot have the same age. That excludes line 10, so the correct answer is line 11:

3 3 8

DFM

1,2,3,4,6,8,9,12,18,24,36,72

Call the ages of the three kids A, B, & C and, without loss of generality, assume that A <= B <= C (It is always possible to arrange it thusly.)

Table below shows all possible combinations.

Line # | A | B | C | A*B*C | A+B+C |

1 | 1 | 1 | 72 | 72 | 74 |

2 | 1 | 2 | 36 | 72 | 39 |

3 | 1 | 3 | 24 | 72 | 28 |

4 | 1 | 4 | 18 | 72 | 23 |

5 | 1 | 6 | 12 | 72 | 19 |

6 | 1 | 8 | 9 | 72 | 18 |

7 | 2 | 2 | 18 | 72 | 22 |

8 | 2 | 3 | 12 | 72 | 17 |

9 | 2 | 4 | 9 | 72 | 15 |

10 | 2 | 6 | 6 | 72 | 14 |

11 | 3 | 3 | 8 | 72 | 14 |

12 | 3 | 4 | 6 | 72 | 13 |

Since knowing the house number is not enough, clearly there are two combinations that have the same total A+B+C, which are the two lines ending in 14, that is, lines 10 and 11. Since there is an oldest child, the top two kids cannot have the same age. That excludes line 10, so the correct answer is line 11:

3 3 8

DFM

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