Courtney Pelowski, Mike Dewitt, Shaun Lewis, Ray Kremer
Scott Peters also submitted a correct solution. Correct solutions
were also submitted by Ron Welch, Yan Fridman, Aaron Kahn, Tim Kelly, Bill
Webb, Cyril Terakopiantz, Steve Prowse, Emanuele Macri, Loranzo Pozzoli,
Matt Humbard, Greg Paul, Philippe Fondanaiche, Dane Brooke.
Since the product of the ages is 72, the ages must be
among the factors of 72. This gives the following possibilitites.
| First Age | Second Age | Third Age | Sum of Ages |
| 1 | 1 | 72 | 74 |
| 1 | 2 | 36 | 39 |
| 1 | 3 | 24 | 28 |
| 1 | 4 | 18 | 23 |
| 1 | 6 | 12 | 19 |
| 1 | 8 | 9 | 18 |
| 2 | 2 | 18 | 22 |
| 2 | 3 | 12 | 17 |
| 2 | 4 | 9 | 15 |
| 2 | 6 | 6 | 14 |
| 3 | 3 | 8 | 14 |
| 3 | 6 | 4 | 13 |
Since knowing the sum of the ages did not uniquely identify the ages of the nieces, you must conclude that the sum of the ages is 14, as no other sum appears more than once. The final information about the likes of the eldest niece tells the neighbor that there is an eldest niece, which would not be the case if the nieces ages were 2, 6 and 6. The only remaining possibility is that the nieces are 3, 3 and 8, and that incidentally, the house number is 14.
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Page last updated 13 January 2000.
©2000 Alberto L. Delgado