I can't figure out this riddle

ClubMan said:
I've seen this problem stated differently and mentioning the house opposite. Are you sure that you have it right above?
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ya - i definitely have it right.

i suppose you could also phrase whereby the sum of the ages is equal to the house accross teh road.

all that matters is that the statistician knows the number the ages add up to.
 
please tell me it's 2, 2, 9 so I can go to bed...

But how do we know that given the information above?
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We don't need to know. We only need to know there is multiple solutions to x+y+z = housenumber
 
leghorn said:
Ah, so we need an answer where the house number can only be added up multiple ways.

2, 2, 9 is the answer.
The sum is 13.

He had to get more info because 1, 6, 6 also works, but there has to be one eldest.
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Well done leghorn - gold star for you !

Yes - for all comboinations of numbers that mutiply to give 36, the sum is unique other than 13.

E.g. 3,3,4 = 36 with a sum of 10.

If the house number was 10,then there would be only 1 possible combination.
Therefore the statistician would have known on teh spot.

However - he says he doesnt have enough infor.

Therefore clearly the house number was 13 as 6,6,1 and 9,2,2, are valid answers.

Whewn he is told the eldest is playing teh piano then it has to be 9,2,2,
 
Ah - got it. You can go to bed leghorn. You have it. There are only three (integral) factors of 36 which add up to the same number 2 x 2 x 9 and 1 x 6 x 6 so to resolve which was correct the statistician needed some more info.

Update: too late - crossed with previous post.
 
If anyone is hungry for more,I do have another one which is by far the toughest i have ever come accross.
Definitely teh daddy of all these riddles !

Anyone who gets this seriously deserves a gold star in my book !

You have 8 balls all of equal size.
7 are the same weight - 1 rogue ball is a different weight.
Note; We dont know if it is heavier or lighter - all we know is that it is a different weight.

You have a balancing scales.

You have 3 weighs to determine which is the rogue ball.

How do you do it

Anyway - i'm off to bed on that note.

By teh way - I can't remember the answer to thsi right now.
I'll have a think about t tomorrow.

I definitely have the question correct though in case anyone is questioning it.
 
I'm assuming that this is a balance where you put say 2 balls on one side and 2 on the other. I'm going to label the two sides of the balance L and R

Measure 1
Take 3 balls (a, b, c) on one L side of the balance and take 3 balls (d, e, f) and put them R side of the balance.
Is L side heavier/lighter/balanced?

IF Balanced, solve for g, h:
If it is balanced then you know that none of these balls are the errant ball.
- Take ball a (now known to be good) and put it on the balance and ball g on the other side.
- If it balances - the errant ball is h, if not then it's g.
- Solved in 2 measures.

Measure 2 [ where un-balanced, so now know that g, h are good]
However if after that first balance it was not even, then take balls (a, b) on L side and (c, d) on R side.
Is L side heavier/lighter/balanced?

IF balanced, solve for e, f
- If it balances - the errant ball is either e, or f
- Put ball e on the balance with g (known good ball). If balanced errant ball is f. Similarly if not balanced, errant ball is e.
- Solved in 3 measures

Step 3 - Note no measure taken here [Now know that e, f, g, h are good] this is where Measure 1 and 2 were not balanced.
If for measure 1 and 2 L side was lighter each time or heavier each time. Then ball c is good - it moved from one side to the other and the direction did not change.
If the L side was lighter in one and heavier in the other (that is it changed) then the errant ball moved sides. Hence c is the errant ball.
Otherwise, we now know balls c, e, f, g, h are good

Measure 3 - solving for one of a, b, d
Put a on L side and b on R side. If balanced then errant ball is d
If L side is still either Lighter or heavier (same direction as previous two measures) then the errant ball has not moved from that side and hence the errant ball is a.
Or if now the balance has swung the opposite direction, then the errant ball has moved and hence the errant ball is b.

Solved in 3!

The insomniac - and no I didn't google or know the answer before. I've had the weighing scales out for the last two hours
 

Excellent.
That's a tough puzzle.

For anyone who is reading the answer and thinks "Ah...it's not that hard", it only seems that way when the answer is written down as conscisely as it was above.

Definite gold star for you dem_shyp.
 
......OK so the bellboy stole the €2 from the €5 and the lads paid €27, no €25 and the hotel manager said €30 but........oh my head hurts....
 
Teabag said:
......OK so the bellboy stole the €2 from the €5 and the lads paid €27, no €25 and the hotel manager said €30 but........oh my head hurts....
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Yep - Blame it on the bellboy !
 
Well for a tenner each, I doubt they'd get their own rooms.
 

Changing the amounts now to nine balls of equal size and one of the balls is slightly heavier and you still can only use the scales three times.

Find the heavy ball!
 
S.L.F said:
Changing the amounts now to nine balls of equal size and one of the balls is slightly heavier and you still can only use the scales three times.

Find the heavy ball!
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well if you know that one is heavier (unlike the earlier puzzle where all you knew was that it was a different weight i.e. could be either heavier or lighter)), then i think it can be done in 2 weighs.

You have 9 balls each numered 1 to 9.

Weigh balls 1,2,3 against 4,5,6.

Depending on the result of that you then know what group of 3 the heavy ball is in.
i.e. if they weigh the same then it's in the group 7,8,9.
If not then it is in the heavier side from the first weigh.

Next - take the group of 3 the heavy ball is - call them a,b,c - and use the same mewthod from the first weigh to determine the heavy ball
 
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