The Gun Counter

Two former college roommates, both logicians, meet at a conference after many years without contact. While catching up, the two eventually get around to discussing their children. The first logician asks the second how many children he has, and what their ages are. The second replies that he has 3 children, but (ever the logician) he will only reveal clues about their ages. The first logician must deduce for himself the ages of the second's children.

"First," says the logician, "the product of my children's ages is 36."

"Second, the sum of their ages is the same as our apartment number in college."

"Third, my oldest child has red hair."

Upon hearing the third clue, the first logician replies at once with the ages of his friend's children. What are they? How do you know?

Cybrludite wrote:Two former college roommates, both logicians, meet at a conference after many years without contact. While catching up, the two eventually get around to discussing their children. The first logician asks the second how many children he has, and what their ages are. The second replies that he has 3 children, but (ever the logician) he will only reveal clues about their ages. The first logician must deduce for himself the ages of the second's children.

"First," says the logician, "the product of my children's ages is 36."

"Second, the sum of their ages is the same as our apartment number in college."

"Third, my oldest child has red hair."

Upon hearing the third clue, the first logician replies at once with the ages of his friend's children. What are they? How do you know?

One problem is that we have to assume whole ages, if we allow for example one kid to be 2-1/2 it kind of mucks the maths up. A lot.
The third clue says there is an oldest, so if there are twins it isn't the oldest one.
The only ages that have an "oldest" would be:

AGES SUM
1x1x36 (38)
1x2x18 (21)
1x3x12 (16)
1x4x9 (14)
2x2x9 (13)
2x3x6 (11)
3x3x4 (10)

Beyond this point I am at a loss because any one of those sums could be an apartment number.

The logicians had a mutual redheaded acquaintance in college, which establishes a time window for conceiving a red-headed child.

evan price wrote:

Cybrludite wrote:Two former college roommates, both logicians, meet at a conference after many years without contact. While catching up, the two eventually get around to discussing their children. The first logician asks the second how many children he has, and what their ages are. The second replies that he has 3 children, but (ever the logician) he will only reveal clues about their ages. The first logician must deduce for himself the ages of the second's children.

"First," says the logician, "the product of my children's ages is 36."

"Second, the sum of their ages is the same as our apartment number in college."

"Third, my oldest child has red hair."

Upon hearing the third clue, the first logician replies at once with the ages of his friend's children. What are they? How do you know?

One problem is that we have to assume whole ages, if we allow for example one kid to be 2-1/2 it kind of mucks the maths up. A lot.
The third clue says there is an oldest, so if there are twins it isn't the oldest one.
The only ages that have an "oldest" would be:

AGES SUM
1x1x36 (38)
1x2x18 (21)
1x3x12 (16)
1x4x9 (14)
1x6x6 (13)
2x2x9 (13)
2x3x6 (11)
3x3x4 (10)

Beyond this point I am at a loss because any one of those sums could be an apartment number.

You missed one.

Durham68 wrote:

evan price wrote:

Cybrludite wrote:Two former college roommates, both logicians, meet at a conference after many years without contact. While catching up, the two eventually get around to discussing their children. The first logician asks the second how many children he has, and what their ages are. The second replies that he has 3 children, but (ever the logician) he will only reveal clues about their ages. The first logician must deduce for himself the ages of the second's children.

"First," says the logician, "the product of my children's ages is 36."

"Second, the sum of their ages is the same as our apartment number in college."

"Third, my oldest child has red hair."

Upon hearing the third clue, the first logician replies at once with the ages of his friend's children. What are they? How do you know?

One problem is that we have to assume whole ages, if we allow for example one kid to be 2-1/2 it kind of mucks the maths up. A lot.
The third clue says there is an oldest, so if there are twins it isn't the oldest one.
The only ages that have an "oldest" would be:

AGES SUM
1x1x36 (38)
1x2x18 (21)
1x3x12 (16)
1x4x9 (14)
1x6x6 (13)
2x2x9 (13)
2x3x6 (11)
3x3x4 (10)

Beyond this point I am at a loss because any one of those sums could be an apartment number.

You missed one.

Nope, because we know the oldest is a singleton, not twins. (It's one of the few things we do know.)

Anyway, this is a pretty lousy puzzle.

Greg wrote:

Durham68 wrote:

evan price wrote: One problem is that we have to assume whole ages, if we allow for example one kid to be 2-1/2 it kind of mucks the maths up. A lot.
The third clue says there is an oldest, so if there are twins it isn't the oldest one.
The only ages that have an "oldest" would be:

AGES SUM
1x1x36 (38)
1x2x18 (21)
1x3x12 (16)
1x4x9 (14)
1x6x6 (13)
2x2x9 (13)
2x3x6 (11)
3x3x4 (10)

Beyond this point I am at a loss because any one of those sums could be an apartment number.

You missed one.

Nope, because we know the oldest is a singleton, not twins. (It's one of the few things we do know.)

Anyway, this is a pretty lousy puzzle.

If we assume the first logician needed the third clue to solve the problem, the answer would be 2-2-9 because 13 is the only sum that comes up twice.

We can't make that assumption because all the clues are presented in sequence with no specific distinction as to how they impact the problem our that they must be needed in an order.

evan price wrote:We can't make that assumption because all the clues are presented in sequence with no specific distinction as to how they impact the problem our that they must be needed in an order.

You do need all three clues, otherwise they logically wouldn't have included them. Take them in the order given.

First clue gives us the list of possibles- Prime factoring of 36 is 2x2x3x3. This gives all possible age combinations as (with sums) would be exactly what Greg listed:

1x1x36 (38)
1x2x18 (21)
1x3x12 (16)
1x4x9 (14)
1x6x6 (13)
2x2x9 (13)
2x3x6 (11)
3x3x4 (10)

This lists all possibilities for the ages, with subsequent clues acting to define subsets of this list of all possibles. Their order is irrelevant to the 1st logician, but he does need all the clues.

Second clue: Sum of ages = Apt #.
The logician already knows the apartment number is 13. We don't. Because he doesn't answer, this means there must be two combinations which sum to the correct apartment number, thus giving:
1x6x6 (13)
2x2x9 (13)

Third clue: There is an oldest.
The moments-older 6 y/o twin would likely not be referred to as 'the oldest' and in the context of a logician riddle, the 9 y/o becomes the obvious correct choice, giving 2,2, and 9 as the ages.

************************************
Had the clues been given in a different order, this would give us the following:
Second clue: There is an oldest.
This eliminates the 1,6,6 combination only, leaving the rest of the list.

Third clue: Sum = apt #.
Logician already knows it's 13, so 2,2,9 is instantly the answer. WE can deduce this as well, since the only thing eliminated by the (alternate) 2nd clue was the other combination which summed to 13.

In either order, the logician needs all 3 clues.