Solution to:
Pastor Petersen
The pastor's age is 49.
Explanation: first, we resolve 2450 into factors: 2450=2*5*5*7*7. These
five
prime factors can be assigned to the three women in the following 20
ways:
| woman 1 | woman 2 | woman 3 |
| 1 | 1 | 2*5*5*7*7 |
| 1 | 2 | 5*5*7*7 |
| 1 | 2*5 | 5*7*7 |
| 1 | 2*5*5 | 7*7 |
| 1 | 2*5*5*7 | 7 |
| 1 | 2*7 | 5*5*7 |
| 1 | 2*5*7 | 5*7 |
| 1 | 2*5*7*7 | 5 |
| 1 | 2*7*7 | 5*5 |
| 2 | 5 | 5*7*7 |
| 2 | 5*5 | 7*7 |
| 2 | 5*5*7 | 7 |
| 2 | 5*7 | 5*7 |
| 2*5 | 5 | 7*7 |
| 2*5 | 5*7 | 7 |
| 2*7 | 5 | 5*7 |
| 2*7 | 5*5 | 7 |
| 2*5*5 | 7 | 7 |
| 2*5*7 | 5 | 7 |
| 2*7*7 | 5 | 5 |
This gives the following possible ages for the three women:
| woman 1 | woman 2 | woman 3 | total age |
| 1 | 1 | 2450 | 2452 |
| 1 | 2 | 1225 | 1228 |
| 1 | 10 | 245 | 256 |
| 1 | 50 | 49 | 100 |
| 1 | 350 | 7 | 358 |
| 1 | 14 | 175 | 190 |
| 1 | 70 | 35 | 106 |
| 1 | 490 | 5 | 496 |
| 1 | 98 | 25 | 124 |
| 2 | 5 | 245 | 252 |
| 2 | 25 | 49 | 72 |
| 2 | 175 | 7 | 76 |
| 2 | 35 | 35 | 184 |
| 10 | 5 | 49 | 64 |
| 10 | 35 | 7 | 52 |
| 14 | 5 | 35 | 54 |
| 14 | 25 | 7 | 46 |
| 50 | 7 | 7 | 64 |
| 70 | 5 | 7 | 82 |
| 98 | 5 | 5 | 108 |
As can be seen, there are two combinations that yield 64 as total age.
All
other totals occur only once. Since the teacher could not conclude how
old each of the three women was based on the product and the sum, we
conclude
that the teacher must be 64. So the possible ages for the women are
5, 10, and 49, or 7, 7, and 50. Now we must use the second hint, which
says
that the teacher can work out how old the three women are if he knows
that
the oldest of the women is older than the pastor. If the pastor would be
younger than 49, the puzzle would still not be solvable, since the
oldest
woman is either 49 or 50. But if the pastor would be older than 49,
the oldest woman would not be older than the pastor. So the pastor's age
is 49.
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