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bonanova

The eldest son

Question

With a nod to jasen's recent and interesting puzzle,

A traveler happened upon a village of huts, laid out as the circles in the  picture below. The village's mayor explained to the traveler that the family living in each of the huts had an eldest son whose age was unique within the village. (No two eldest sons had the same age.) How interesting, replied the traveler. Tell me this: of all the male children here, what is the age of the very oldest?

The mayor thought for a moment and replied, well I guess I could tell you that none of them are yet of voting age (21), and I guess you might be interested to hear that there are no gaps in their collective ages. But all of that wouldn't be enough information. I think it would be better for you to just knock on all the doors and ask.

nineteen.gif

I don't have time for that, replied the traveler, and I'm really not that interested. Well, here's an interesting thing about our village, replied the mayor. You may have noticed, our huts are laid out so that many rows of 3, 4, or 5 huts cut across the entire village. Just ask at the huts along any of those rows. Add the ages that you hear, and divide the sum by two. That way you will learn the age of the oldest son in the village.

And now, without knocking on any doors, you can learn it too.

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10 answers to this question

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Spoiler

There are 3+4+5+4+3 = 19 huts so 19 ages of the eldest sons in each hut.
Sum of all ages is 5 times the sum of any row which is twice the age of the oldest sun in the village.
There are no gaps and no two are the same so the sum of the 19 ages.
Hence:
(i) S = 10*X
(ii) S = Y + (Y+1) + .... +(Y+18)
(iii) X = Y+18

Rewriting (ii) as a consecutive sum:
S = 19*(Y-1) + (1+....+19) = 19*(Y-1) + 19*20/2
and using (i) and (iii):
10*X = 10*(Y-1)+10*19 = S = 19*(Y-1) + 19*10

Therefore 9*(Y-1) = 0, so Y=1 and X = 19.

Did not use the fact that they are not of voting age (which indeed narrows the possibilities for Y to be either 1 or 2).

 

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On ‎10‎/‎22‎/‎2016 at 5:20 PM, bonanova said:

With a nod to jasen's recent and interesting puzzle,

A traveler happened upon a village of huts, laid out as the circles in the  picture below. The village's mayor explained to the traveler that the family living in each of the huts had an eldest son whose age was unique within the village. (No two eldest sons had the same age.) How interesting, replied the traveler. Tell me this: of all the male children here, what is the age of the very oldest?

The mayor thought for a moment and replied, well I guess I could tell you that none of them are yet of voting age (21), and I guess you might be interested to hear that there are no gaps in their collective ages. But all of that wouldn't be enough information. I think it would be better for you to just knock on all the doors and ask.

nineteen.gif

I don't have time for that, replied the traveler, and I'm really not that interested. Well, here's an interesting thing about our village, replied the mayor. You may have noticed, our huts are laid out so that many rows of 3, 4, or 5 huts cut across the entire village. Just ask at the huts along any of those rows. Add the ages that you hear, and divide the sum by two. That way you will learn the age of the oldest son in the village.

And now, without knocking on any doors, you can learn it too.

what is meant by " there are no gaps in their collective ages"? Does it mean that the total age of all the male children in every hut is the same number?

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On ‎10‎/‎23‎/‎2016 at 4:12 PM, araver said:
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There are 3+4+5+4+3 = 19 huts so 19 ages of the eldest sons in each hut.
Sum of all ages is 5 times the sum of any row which is twice the age of the oldest sun in the village.
There are no gaps and no two are the same so the sum of the 19 ages.
Hence:
(i) S = 10*X
(ii) S = Y + (Y+1) + .... +(Y+18)
(iii) X = Y+18

Rewriting (ii) as a consecutive sum:
S = 19*(Y-1) + (1+....+19) = 19*(Y-1) + 19*20/2
and using (i) and (iii):
10*X = 10*(Y-1)+10*19 = S = 19*(Y-1) + 19*10

Therefore 9*(Y-1) = 0, so Y=1 and X = 19.

Did not use the fact that they are not of voting age (which indeed narrows the possibilities for Y to be either 1 or 2).

 

when you say the sum of all ages do you mean the sum of the ages of all the eldest sons or sum of the ages of all the male children in the village?

what is Y?

I really like the puzzle.  But I am neither able to completely understand the question or the solution.  Can you throw some light on both?

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@FUZZYI think he means if you list out their ages in order, there will be no repetition and no gaps. So ages 1-19, 2-20, 3-21, etc.

Sum of ages, I would assume applies to eldest sons, not all males.

In araver's solution, S is the sum of ages, X is the oldest son in the village, and Y appears to be the youngest.

I'm not sure where the y-1 came from. If you don't get that either, here's another way of looking at it using the same 3 original equations. Probably just a slightly longer form of his solution.

Spoiler

(i) S = 10*X
(ii) S = Y + (Y+1) + .... +(Y+18)
(iii) X = Y+18

Rewrite (ii) with "y"s combined. S = 19Y + (1 + 2 + . . . + 18) = 19Y + 171

Set equal to (i). 19Y + 171 = 10X

Substitute Y + 18 for X (iii). 19Y + 171 = 10(Y + 18)

Distribute and solve for Y. 19Y +171 = 10Y + 180  ===>  9Y = 9  ===> Y=1

Solve for X using (iii). X = 1 + 18 = 19

@araver Maybe a typo. 10*(Y-1)+10*19 = S = 19*(Y-1) + 19*10

The "10*19"s cancel along with the "(y-1)"s leaving 10=19?

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17 hours ago, Thalia said:

 

In araver's solution, S is the sum of ages, X is the oldest son in the village, and Y appears to be the youngest.

I'm not sure where the y-1 came from.

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(i) S = 10*X
(ii) S = Y + (Y+1) + .... +(Y+18)
(iii) X = Y+18

Rewrite (ii) with "y"s combined. S = 19Y + (1 + 2 + . . . 

@araver Maybe a typo. 10*(Y-1)+10*19 = S = 19*(Y-1) + 19*10

The "10*19"s cancel along with the "(y-1)"s leaving 10=19?

@Thalia nope that was not a typo. One cannot divide (y-1) because it is 0, hence the solution

I believe it looks weird because I created the terms on purpose to cancel out 19*10 from the equation instead of solving it as a*y + b = 0. Aka substitution Z=Y-1 leads to a simpler equation with no free term.

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23 hours ago, Thalia said:

@FUZZYI think he means if you list out their ages in order, there will be no repetition and no gaps. So ages 1-19, 2-20, 3-21, etc.

Sum of ages, I would assume applies to eldest sons, not all males.

In araver's solution, S is the sum of ages, X is the oldest son in the village, and Y appears to be the youngest.

I'm not sure where the y-1 came from. If you don't get that either, here's another way of looking at it using the same 3 original equations. Probably just a slightly longer form of his solution.

Thank you Thalia.  Your explanation made both the Q and A very clear.

 

 

 

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19 hours ago, araver said:

@Thalia nope that was not a typo. One cannot divide (y-1) because it is 0, hence the solution

I believe it looks weird because I created the terms on purpose to cancel out 19*10 from the equation instead of solving it as a*y + b = 0. Aka substitution Z=Y-1 leads to a simpler equation with no free term.

Ooh. Dividing by 0. My physics teacher did that once. Still not sure how. I can see how y-1=0 knowing the solution already. But how is that known to be 0 before the solution? What made you pick y-1 as opposed to anything else?

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25 minutes ago, Thalia said:

Ooh. Dividing by 0. My physics teacher did that once. Still not sure how. I can see how y-1=0 knowing the solution already. But how is that known to be 0 before the solution? What made you pick y-1 as opposed to anything else?

I just saw that the terms that did not contain y are very close and 10 and 19 can be rewritten to reduce the terms. Summing 1+2+...+N is easier and gives 19*20/2 which is 19*10 and 10 was the multiplier in the other equation.

Hmm, it's hard to explain how the mind works. It just felt right at that moment to go on that road. Like a piece of a puzzle sliding in.

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