A doctor in Michigan has a sister who lives in California but the sister does not have any brother that lives in Michigan, how come??
:P

*twins there = two eldest

republic, I think i have the solution, its just in my head i have not tried on paper yet, correct me if I am wrong.

soln,
you leave one stack, weigh only 9 stacks but with following order
1st stack -- 1 coin
2nd stack -- 2 coins
3rd stack -- 3 coins
4th -- 4
5th -- 5
6th -- 6
7th -- 7
8th -- 8
9th -- 9

if it was all 100 gm coins it would weight (x)=900+800+700+600+500+400+300+200+100

When we weight if the weight is equal to x, then the odd coin is from the stack that we left out.
if the weight is 10 gram less its the 1st stack
20 gm less -- 2nd stack
30 gm less -- 3rd stack
40 gm less -- 4th stack
50 gm less -- 5th stack
60 gm less -- 6yh stack
70 gm less -- 7th stack
80 gm less -- 8th stack
90 gm less -- 9th stack

I think thats the answer ....... damn it was a good one. thank you republic keep it coming.

Timetraveller,

Guest's total 14 logic makes sense as that is the ony possibility for the mathematician to get confused as she already knows the number of windows ...... as opposed to ur 3 solutions but your logic of red hair makes sense than guest's as mathematician just wanted to make sure wheather there is elder or not, didnot want to know abt the red hair in fact.

Here is a new question.

A farmer called his five sons and distributed the some oranges among them as below:
1st: 100 oranges
2nd: 80
3rd: 60
4th: 40
5th: 30

Then the farmer asked his son to go to five different places to sell the oranges.
The conditions set are:
1. They can not redistribute the oranges.
2. All oranges should be sold at same rate.
3. Final collection should be equal for each.

Now what is the rate offered and final collection?

Ah! Interesting question - answer going on here...Before, I look into other questions and solutions, I think I would work on slow poison's question.

A question, though--you mean, that they all can sell as many oranges as they want, but the price has to be same, right?

Not as many. The number is fixed.
For first son only 100
secod: only 80
third: 60
fourth: 40
and last one: 30

no more no less

same rate

same sum

No, what I mean is they have got the different number of oranges, but they can sell as much as they want, right?

If not, how can you sell 100 oranges at the same rate to 80 oranges and get the total amount equal? :S

They can not sell what ever they want. They have to sell all.

Seems impossible! But not.

They didn't sell any. That is the only mathmatical solution I can come up with. :)

Or, the rate/orange they gave was $0.00 and sold ALL! :D

I was not expecting this answer

Seems I have to give a Hint.

They sold all oranges in two shift but with same rate in each shift.

the orange is simple
lets say if A sold each orange in one rupee then he got 100 rupees,
then B sold his oranges at 1.25 each and collected 100 rupees
similarly C sold at lets say 1.66 per orange and collected 100 rupees

and so on

or is it that i misunderstood the question

Is it this way?

lets only consider two sons, with 100 and 80 oranges.


A sold 96 atr 0.5 and got 48rs. And 4 atr 1 to make 4. total 52rs.
B sold 56 atr 0.5 to make 28rs. 24 atr 1 to make 24. total 52.

maranchyase got the answer i think, what is the answer slowpoison ..........

oops.

my apologies.

gues 4 is right. there are two different sums. and several younger twins possibility.

sorry.

but yeah, that is the answer.

Maranchyse answer won't work for 40 and 30 case. The the idea is rite.

This is my answer

First shift: 11 oranges @ 1 Rs

person no. sold money collected no. remained

A(100) 99 9 1
B (80) 77 7 3
C (60) 55 5 5
D (40) 33 3 7
E(30) 22 2 8

Second Shift: 1 orange @ 1 Rs

person sold total money

A 1 9+1=10
B 3 7+3=10
C 5 5+5=10
D 7 3+7=10
E 8 2+8=10

This idea works for any number multiple of 10 and less than or equal to 100.

Sory the tabular format is changed so calculete with the rate given.

A ten-volume encyclopedia is arranged on a bookshelf - each volume has 1000 pages. If a bookworm starts eating the pages from the first page of volume 1 and stops when it eats the last page of volume 10, how many pages does it eat? Assume it eats all the pages in order and ignores the covers.

Doctor can also be famale.

Republican! I guess the answer is
1+8*1000+1
=8002