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Conference rusure::math

Title:	Mathematics at DEC

Moderator:	RUSURE::EDP

Created:	Mon Feb 03 1986
Last Modified:	Fri Jun 06 1997
Last Successful Update:	Fri Jun 06 1997
Number of topics:	2083
Total number of notes:	14613

217.0. "Mr. E's daughters ages" by SPRITE::OSMAN () Fri Feb 01 1985 15:54

This is a simpler problem of the ilk of 110.0.

The mailman comments on the attrictiveness of Mr. E's three daughters.
(For those of you bothered by my use of mailMAN instead of mailPERSON,
please note that Mr. E is home to receive the mail because his wife
works every day !).

Mailman:	How old are they ?

"Well", explains Mr. E, "their ages add to the number on my front door".

Mailman:	Hmmm.  I still don't know their ages !  I need more 
		information.

Mr. E:		Well, their ages multiply to 36.

Mailman:	Sorry, I still don't know !

Mr. E:		My youngest daughter prefers Cherry pie to Apple.

Mailman:	Aha !  Now I know.  Their ages are . . .

[your turns, oh great math notes file readers]

T.R	Title	User	Date	Lines
217.1		CASTOR::MCCARTHY	`Sun Feb 03 1985 18:37`	12
	The ages are 3, 4, and 6. Since the ages multiply to 36, we have three choices: (3, 3, and 4) (3, 4, and 6) and (2, 2, and 9) Since the answer is ambiguous knowing the sum, this leaves (3, 4, and 6) and (2, 2, and 9) which both sum 13. Since there is a youngest daughter, the second choice (which involves twins) is eliminated.
217.2		HARE::STAN	`Sun Feb 03 1985 22:23`	1
	Why can't one kid be 1 year old?
217.3		TURTLE::GILBERT	`Mon Feb 04 1985 02:28`	3
	Besides which, 3x4x6 does not equal 36. (correct answer left as an excercise).
217.4		SPRITE::OSMAN	`Mon Feb 04 1985 10:55`	7
	I continue to leave the answer as an exercise. However, for those enjoying the harder variety, I've started to wonder the following: If we change the third clue from "their ages multiply to 36" to "their ages multiply to the postman's age", do we have a solution ? If not, can we introduce some reasonable upper limit on the oldest daughter's age such that we do have a solution ?
217.5		CASTOR::MCCARTHY	`Mon Feb 04 1985 16:41`	11
	Not only can, but has to be. My previous reply was pretty obvious operator headspace, since 346 is 72, not 36. The control store has always been more reliable than the ALU. The correct answer seems to be: out of (1,1,36), (1,2,18), (1,3,12), (1,4,9), (1,6,6), (2,2,9), (2,3,6) and (3,3,4). The ambiguous cases are (1,6,6) and (2,2,9). Therefore the correct answer seems to be: (,11,6,6)
217.6		CASTOR::MCCARTHY	`Mon Feb 04 1985 16:43`	1
	re: .-1 one childs age must be one.
217.7		CASTOR::MCCARTHY	`Tue Feb 05 1985 15:40`	12
	Assuming retirement age for postmen is 70 or thereabout, we don't need to know the postman's age, merely that it is the product of the three ages. The next number after 36 which has two different sets of three factors with the same sum (and which is made unambiguous by the third condition (my youngest daughter...)) is 90. (10,3,3) and (9,5,2) (This time I checked the arithmetic). Rasing the life expectency of postmen to 500 years increases the number of possibilities a lot. (Haven't checked them all yet). Limiting the eldest daughter's age doesn't help much. the combinations in the 500 range all involve sets of factors less than 20.
217.8		SPRITE::OSMAN	`Wed Feb 06 1985 10:03`	2
	What about products LESS than 36 ? Are there any numbers LESS than 36 having two or more triplets whose sum are the same ?
217.9		CASTOR::MCCARTHY	`Wed Feb 06 1985 19:54`	7
	None that I could find. The results were obtained with a program that generates all combinations of at least three prime factors with a product less than 100. For each combination, all possible sets of three factors were generated. These are scanned for matching sums. It found two results in the 0-100 range. I upped the limit to 500 and it found more. That was only using primes below 100, however. Using more primes should generate more results.
217.10		CASTOR::MCCARTHY	`Wed Feb 06 1985 19:55`	2
	Although I believe not many, since the solutions all seem to involve small primes (2,3,5).
217.11		SPRITE::OSMAN	`Mon Feb 11 1985 14:48`	22
	So, final result seems to be that we now have a "better" problem, i.e. more amazing to the casual reader that it can possibly be solvable, since it contains few numbers and seems to not have enough information (although not as amazing as 110.0 !): Mailman: What are the ages of your three daughters ? Mr. E : They add to the house number. Mailman: Even knowing your house number, I still need more information. Mr. E : They multiply to your age ! Mailman: Hmmm. Even knowing my own age isn't enough ! Mr. E : The youngest likes cherry pie. Mailman: Aha ! Now I know. Well, the reader needs to know that the mailman is below retirement age of 65. How old are the three daughters ?
217.12		AUSSIE::GARSON	`Fri Feb 26 1993 23:13`	5
	Mr E's eldest daughter is now grown up and has done her BSc in mathematical sciences. She has inherited the family trait of being a bit coy about ages but she does volunteer the following information. The remainder on dividing her age by the numbers x, y and z is r, s and t respectively. (She tells you the six numbers.) Find a formula for her age.