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

that was a very good one chyamekancha ........ kept me laughing ...... swatantratagamy first draw the whole picture who lives in which house and you will know who owns fish ........ I will tell you the answer let others try too ...... again ice one chyamekancha

my question was just for fun, if you loved it its my pleasure

i am sure miss Y would soon leave her apartment and move down to sixth floor rather than carrying a stick with her. or even further down so that she wouldnot have to jump.
But i wonder why Miss Y lived in that 12th floor, this Miss Y is really confusing me

keep rocking........

simple...

doctor thinks her as sister but she doesnt.....

Mathew that has been answered way before ...... the doctor is lady doctor

Ritthe, I think I got it. And, evidently it must be my last post--the record in my whole Sajha--for it says, I have posted 19 times..Oh, well..

At first it was very complicated, I made chart and when I found I could put something in two or three boxes, It didn't look that complicated, and it became kinda fun for few minutes, then suddenly it became complicated...and this kept going on. The major difficulty was when I messed up in few last steps. Here's it anyway: :D

Norway
Yellow, Dunhill, Water, Cat

Denmark
Horse, Blue, Tea, Blends

Britain
Milk, Red, Pall Mall, Bird

German
Green, Coffee, Prince, FISH

Sweden
White, Beer, Blue, Cat

View/Share this post only

"And, evidently it must be my last post--the record in my whole Sajha"

to be read:

And, evidently it must be my last post--the record in my whole Sajha life"

And, I can post more now, the numbers of posts reduced as the hours passed, I thought that would be the last! :)

So, what you say on my answer, Ritthe?

yeah you got it slackdemic ........ If you understand the clues with the blend one ..... its easy afterwards ..... To me that was the very important clue as it was the clue with age and veteran in the puzzle u posted. it may be my last post too ..... have already been frozen in sajha.com so I tried from gbnc.org ..... it may give me few chances to post ...... goodbye anyway will come up with more good ones and hope u will do the same bye all

Coins puzzle:

You have 10 stacks of coins, each stack containing 10 coins. The coins in 9 of the stacks weigh 100 grams each. The coins in one of the stacks weigh 90 grams each. All 100 coins look identical. Using a weighing scale, can you determine which stack has the lighter coins, if you are allowed to measure the weight only once?

This was the problem I posed a day ago:

Ok, here's a problem my math professor gave me during coffee. A few of us used to hang out and he'd give us these problems.

So here it is. If you have seen it please do not answer unless someone does.

Two ladies met several years after they had graduated from university. One lady- A, was a mathematician. The other lady- B, was a housewife.

B, the housewife, has three sons- x, y, and z. After a few minutes of talking B decided to give a achallenge and asked her to find the age of her three sons. She gave her two hints:

1. x*y*z=72 [i.e. the product of their ages is 72]
2. x+y+z=number of windows in building across the street.

A worked on the problem for a while and finally said, "Can you give me another hint?"
B said "Yes"

A asked "Does your oldest son have red hair?"
B gave the answer and A gave the solution for x, y and z.

Question: Find x, y and z.

A lot of you may dispute the solution in the end.

The solution is 3 3 8

Here's why:

These are the combiations of numbers that make up 72 as a product:
1 2 36
1 3 24
1 4 18
1 6 12
1 8 9
2 2 18
2 4 9
2 6 6
3 4 6
3 3 8

Now, as we can see, any combination can be a solution. So the reason why the mathematician asks is the oldest or eldest son has red hair is because if there is a solution, as posed by the housewife, it must be UNIQUE TO THIS CASE ONLY. Meaning there are so many combinations possible that there must be a unique case that'll determine the answer. In ths case, we see there are three:

2 2 18
2 6 6 and
3 3 8

Now the problem is finding which one of these is UNIQUE that causes this problem to have a solution.

"does your eldest son have red hair?"...and she gives the answer.

The answer the mathematician is looking for is NOT whther the son has RED hair or not. We can see that we could either have 2 6 6- two eldest sons or 2 2 18 and 3 3 8- ONE eldest son in the last case.

So the mathematician is trying to determine if there is an ELDEST son or not. the housewife has been tricked. If she had an eldest son and he did in fact have red hair, she'd say "YES, he does have Red hair" and the solution would be - 3 3 8 or 2 2 18, because there are two younger twins in both cases.

Had she said "No, he does not have red hair", she would still have determined the solution to be 3 3 8 or 2 2 18 because there still exists an ELDEST son.

However, because I said the solution is UNIQUE, there cannot be a posibility that 2 6 6 be an aswer.

Now is it 2 2 18 or 3 3 8? the answer again lies in the nature of the problem- the claim that it is solvable. Because the age 18 appears above as 1 4 18, the housewife should have carefully thought of the problem and given the mathematician these dual scenarios to dabble with.

therefore the answer is 3 3 8.

This why I said this is gonna cause some denials. When I solved this this way, the prof liked it but other students hated it.

lol.

I wont argue any ther reasonings here.

Thanks, this thread is awesome!

Before I give my own reasoning, I am not fully convinced by the reasons you gave. How did you narrow down your "unique" solution to only three possible choices. You did not give any reason! In other words, what makes them unique??

Also, when you eliminated your last answer (of the remaining two) saying 18 appears above, then 2 also appears above. I don't get your reasoning.
--------------------------
Here's what I would do:
To begin with, there are 11 possible combinations.

1+2+36=39
1+3+24=28
1+4+18=23
2+2+18=22
1+6+12=19
1+8+9=18
2+3+12=17
2+4+9=15
2+6+6=14
3+3+8=14
3+4+6=13

In all of the cases above, the product is 72, but their sums are different except for 2, 6, 6 and 3, 3, and 8. The number of windows the mathematician sees is 14. That's why he needs one more hint to know which one among the two is the answer. The number of windows has to be 14 because, if there was another number, say 18, the mathematician would know the answer right away, and it would be: 1, 8, and 9. So, because the sum of their ages (the number of windows) was 14, the mathematician asks for one more hint to which the housewife says: My eldest son has red hair.

If the eldest son has a red hair, then the answer cannot be 2, 6, 6 because then you would have twins there. So, the answer must be 3, 3, 8.
--------------

*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