EInsteins Riddle, try to solve it

[Pameff]

There are five houses with five different colors. In each house lives one person, each person has a different nationality and a distinct favorite drink, cigarette brand, and pet:

1. The owner of the yellow house smokes Dunhill cigarettes.

2. The owner of the middle house drinks milk.

3. The owner of the red house is British.

4. The man owning a horse lives next to the guy smoking Dunhill cigarettes.

5. The person in the first house is Norwegian.

6. The green house is directly left to the white house.

7. The man smoking Winfield likes beer.

8. The man smoking Marlboro lives next to a man owning a cat.

9. The Danish man loves tea.

10. The Norwegian lives next to the blue house.

11. The man smoking Marlboro has a neighbor who only drinks water.

12. The German smokes the Rothmans brand.

13. The owner of the green house drinks coffee.

14. The Swedish guy owns a dog.

15. The person smoking Pall Mall owns a bird.

Now then, who own a fish?

Searching for more interesting riddles ? Visit URL removed by moderator. Free contest starting 1 of June with brandnew and hard to solve riddles..

Edited May 19, 2014 by CaptainPanic
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[J.C.MacSwell]

 

The German

 

[Hendrick Laursen]

An easy riddle,

The official answer could be looked up in Wikipaedia.

Have fun

Edited September 17, 2014 by Hendrick Laursen

[studiot]

 

Searching for more interesting riddles

 

This problem kept me quiet for a couple of hours with a paper and pencil, the first time I did it

 

Fill in the missing digits marked X

Note that none of the Xs are a 7 and that the substituting digits must appear in the places shown.

 

 

 

Since folks are just going to look up answers I will not attribute it for the moment.

Edited September 17, 2014 by studiot

[Phi for All]

An easy riddle,

The official answer could be looked up in Wikipaedia.

Have fun

 

Easy isn't always the most fun. Trying to solve riddles without cheating enhances our intellectual honesty, imo.

[imatfaal]

Hmm - second try (I deleted the first one as it had a second seven - one of the benefits of modhood is the power to erase one's mistakes)

 

 

 

 

 

 

Not sure it is the same pattern you were looking for - but the fact that once I looked only two numbers worked and only one had no other sevens makes me more confident

[Hendrick Laursen]

To cheat or not to cheat

Well, I tried to provide a source for the ones who wanted to check their answer, and the unlucky ones who couldn't find it.

[studiot]

I'm sorry imatfaal, but your solution is not valid.

 

Third time lucky?

 

As a matter of interest when I asked some engineers awhile back they tried a brute force and ignorance method (let the computer try every combination and filter).

One reported 164 sets of matching numbers

another reported 150

 

but neither picked out the one and only correct one.

 

other wrong possibilities include

 

768 * 142 = 109056

762 * 132 = 100584

Edited September 17, 2014 by studiot

[Spyman]

 

100678/142=709

 

 

EDIT - This also seems to fit:

 

 

100536/142=708

 

 

EDIT 2 - Just noticed the rule: "none of the Xs are a 7", which invalidates my first option.

Edited September 18, 2014 by Spyman

[studiot]

Spyman did you have a method or did you just guess?

 

Since this is a science forum it would be nice to also discuss strategies for solving these things (both Einstein's and mine).

Edited September 18, 2014 by studiot

[Spyman]

Well, I looked at imatfaal's second try and then went on from there, I will try to explain my method under the spoiler.

 

 

 

First I noticed that the second digit in the number 7XX had to be a zero to fulfill the whole division with only two subtractions.

(The last subtraction contains a four digit number and would only have had three digits unless the second digit above was a zero.)

 

So we have the possibilities of 701, 702, 703, 704, 705, 706, 707, 708 and 709 for the quotient.

 

Secondly the denominator multiplied with 7 must be a three digit result which limits its maximum to 142.

(143*7=1001 -> to high since the first subtraction contains a three digit number, but 142*7=994 -> closest below limit.)

 

Also the denominator multiplied with the whole quotient must be at least 100000 which limits its minimum to 142.

(141*709=99969 -> to low since numerator contains a six digit number, but 142*709=100678 -> above limit with the highest quotient.)

 

At this point the denominator is known, since max and min limits are the same it must be 142.

 

Since 142*709 fits, I thought that I had found the right answer and made my first post, however later I noticed that I had failed to calculated the minimum limit of the quotient, which turned out to be 708 and found that there was a second answer.

(142*7=994 -> to low since the last subtraction contains a four digit number it must be higher, but 142*8=1136 and 142*9=1278.)

 

After posting my second answer I started thinking that there should only be one answer and found the "no sevens" rule.

(The first answer contains three X as 7 so the second answer was the correct one.)

 

 

[imatfaal]

I'm sorry imatfaal, but your solution is not valid.

 

Third time lucky?

 

As a matter of interest when I asked some engineers awhile back they tried a brute force and ignorance method (let the computer try every combination and filter).

One reported 164 sets of matching numbers

another reported 150

 

but neither picked out the one and only correct one.

 

other wrong possibilities include

 

768 * 142 = 109056

762 * 132 = 100584

 

 

I taught myself LD and would be quite happy with my format (I will happily fill in two digits in one go if I can see the multiple) - but with a bit of thought I see what you mean and why I am wrong. I grew up during that short period when calculators were new and interesting; educators were happy not too teach LM and LD - which, now I am older, I see is complete madness

[studiot]

So here is my method for my problem for comparison.

 

 

OK we are dividing a six digit number ABCDEF by a three digit number, GHI to get a quotient JKL

We are told J=7

Proceeding through the long division:

GHI into ABC won't go so bring down D and move one place right.

GHI into ABCD goes 7 times.

We are told the product 7 times GHI is only three digits ie not more than than 999.

If G is 2 or greater then the product would have four digits.

So G=1

This also tells us that A = 1 since the maximum product 1HI times 7HL is 199 times 799 = 159001 and we don't show leading zeros.

So place 7 times 1HI under the 1BCDEF and do the subtraction.

We are told this results in a two digit remainder so bring down E to make three digits.

We are told this is not divisible by 1HI.

So we place a zero in the quotient to note this fact

This means K=0

Then we bring down the last digit, L to make a four digit number.

We are told that 1HI divides this exactly with no remainder so this four digit number is equal to L times 1HI.

This should tell us something about L.

 

Edited September 18, 2014 by studiot