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Listing all possible cases

Hey I have worked on this problem:

A Minister and a Bishop were having a cup of tea. There was a knock at the door, and three bell ringers entered the room. After introductions, the Bishop asked the Minister how old the bell ringers were. “Well,” the Minister said, knowing the Bishop had a penchant for numerical puzzles, “if you multiplied their three ages together, you’d get 2,450. But if you added them, you’d get twice your age.” “Hmm,” the Bishop muttered, after several moments’ thought. “I haven’t enough information to solve that.” “It may help, my dear Bishop,” offered the Minister, “to know that I am older than anyone else here in the room.” “Yes, indeed it would,” replied the Bishop. “Now I know their ages.” The question is: How old is the Minister? You may assume that all ages are integers.

I found the prime factorisation of 2450 and then made a table and tried to list all the possible values of a,b and c. However I missed some cases.

With these kinds of problemsI always seem to miss some cases... What is the best way to list all cases systematically? I use usually use a table, but most of the time I am not sure how to start listing the cases.

Reply 1

Joinedup

Original post by FXLander

the prime factorisation is 2,5,5,7,7

well since it's peoples ages I think you can safely assume there aren't any 120+ year olds capable of ringing bells in the world...If the riddler wanted to be awkward they'd have made it numbers of stamps in stamp collections or something else that provides no clue.

so there's only 3 combinations where one of the ringers ages will be a product of 3 primes and 2 will always have to be one of those 3 primes.

then you just need to list the combinations where you've got 2 ringers with ages that are the product of 2 primes.

so 9 cases in total