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. Therefore, 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 is. If the pastor would be younger than 49, the puzzle would still not be solvable, since the oldest woman is either 49 or 50. However, if the pastor would be older than 49, the oldest woman would not be older than the pastor is. Therefore, the pastor's age is 49.
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