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  #1  
Old 08-24-2005, 10:04 PM
bdk3clash bdk3clash is offline
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Default A puzzle from Popular Science April 1960 issue

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I totally don't get it. If anyone knows the answer, post it in invisible type I guess.
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  #2  
Old 08-24-2005, 10:13 PM
Glenn Glenn is offline
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Default Re: A puzzle from Popular Science April 1960 issue

<font color="white"> 3 3 25, there are two combinations that add up to 31, and one of the two is 15 15 1, making "are you the eldest" a relevant question </font>

edited for further clarity:

<font color="white"> The way you solve this problem is to start by taking the prime factorization for 225 (3*3*5*5), then find all combinations of 3 factors of 225 using this info (and not forgetting to include 1--such as 1*9*25). Next, take the sum of each set of factors and find a common sum that would require further information.</font>
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  #3  
Old 08-24-2005, 10:16 PM
partygirluk partygirluk is offline
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Default Re: A puzzle from Popular Science April 1960 issue

x * y *z = 225

what are the possible values of x, y and z?

5, 5, 9

15, 15, 1

5, 15, 3

25, 3, 3

25, 9, 1

45, 5, 1

75, 3, 1

The house number is 31. The guy then knows the 3 are either 15, 15, 1 or 25, 3, 3

if it was 15, 15, 1 there would be no "eldest" so it is 15, 15, 1.
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  #4  
Old 08-24-2005, 10:20 PM
bdk3clash bdk3clash is offline
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Default Re: A puzzle from Popular Science April 1960 issue

<font color="white">Why can't they be 1, 9, and 25? Where does 31 come from?</font>
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  #5  
Old 08-24-2005, 10:20 PM
Punker Punker is offline
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Default Re: A puzzle from Popular Science April 1960 issue

Answer in white:

<font color="white"> Options of 3 numbers whose product is 225:

1,1,225 (sum 227)
1,5,45 (sum 51)
1,15,15 (sum 31)
5,5,9 (sum 19)
5,3,15 (sum 23)
25,3,3 (sum 31)

If the house number was any of the unique solutions, the census taker would have the answer (say, if they lived in house #23, he'd know it was 3,5,15). So, since he still doesn't know the answer, the house # must be one of the non unique numbers. That leaves:

1,15,15 (sum 31)
25,3,3 (sum 31)

The respondent could not say he was older if he was one of the two 15 year olds, so that eliminates the 1,15,15 combination. That leaves the 3,3,25 combination and thats the ages.

</font>

See? Easy.
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  #6  
Old 08-24-2005, 10:20 PM
Eurotrash Eurotrash is offline
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Default Re: A puzzle from Popular Science April 1960 issue

edit: nevermind
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  #7  
Old 08-24-2005, 10:21 PM
Punker Punker is offline
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Default Re: A puzzle from Popular Science April 1960 issue

Its process of elimination; it has to be 31 or "the product of our ages is 225 and the sum of our ages is the house number" would be sufficient to solve the problem.
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  #8  
Old 08-24-2005, 10:23 PM
Thythe Thythe is offline
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Default Re: A puzzle from Popular Science April 1960 issue

edit: nevermind...I think 3 of us asked the same question simultaneously.
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  #9  
Old 08-24-2005, 10:24 PM
bdk3clash bdk3clash is offline
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Default Re: A puzzle from Popular Science April 1960 issue

EDIT: This was much easier than figuring it out for myself. You guys are smart!
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  #10  
Old 08-24-2005, 10:25 PM
ddubois ddubois is offline
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Default Re: A puzzle from Popular Science April 1960 issue

x * y * z = 225

225 breaks down into 5x5x3x3, which leaves you a limited number of ways to formulate ages. x, y, and z, must either be one of: 1, 3, 5, 9, 15, 25, 45, 75, and some of those are mutually exclusive (if there's a 45 year old, then the other two people have to be 5 and 1).

Just looking at it for awhile, I notice that 25+3+3 = 31 and
15+15+1 = 31. I'm guessing the "are you the eldest" question is there to eliminate the 15,15,1 case, leaving the 25,3,3 case.
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