Two Brain Teasers

by sophlightning305 on Sunday, February 21, 2010

I think they're hard and haven't read the answers so please think about it with me:


Brain Teaser #1: The King and His Coins (Difficulty Level: Easy)

The king of the land has 12 districts that he rules over. At the end of each month, each district is responsible for paying taxes. Taxes are 100 gold pieces - each weighing one once. Lately, there has been rumblings in the kingdom and the King thinks that one of his districts could be short changing him but he is not sure which one. Specifically, the king suspects that one district is turning in coins that look like the one ounce coins but in fact only way .9 of one ounce. The King has a digital scale (gives you the weight of whatever is on it instead of telling you which side is heavier) that gives the weight of objects in ounces but it can only be used for one measurement and then it breaks. As a result, the King is stumped and doesn’t know how to find the culprit – that’s when he requests your services. You are the town wiseman. He thinks that you will be able to tell him which district is cheating him. However the scale can be used for only one measuring. How would you do it?

Brain Teaser #2: The Oldest Child (Difficulty Level: Difficult)

Two men that haven’t seen each other in many year ran into each other in the street. The first man says to the second that it’s great to see him after so many years and asks weather he has a family or not. The second man replies that he does have a family and that it consists of a wife and three children. The first man then inquires about the ages of the three children. The second man answers this question in the form of a riddle. He says that the product of his three children’s ages is equal to 36 and that the sum of the their three ages is equal to the number of the blue house on the corner of the road. Further, all of the ages are exact unit numbers and no fractional equivalents should be used. The first man walks up to the blue house on the corner, looks at the number, and returns to the second man. The first man then states that the second did not give him enough information to solve the riddle. The second man replies that the first man is correct and that one additional clue is needed to solve the riddle. The second man smiles and says that his oldest child’s eyes are blue. The candidates task is to tell you the ages of the three children and correctly identify the number on the house.


The typos are not mine...they were like this in the packet lol.

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11 comments:

Comment by Martias on February 22, 2010 at 9:46 AM

For the first one: Luckily for you, multiples of nine are easy. Just put one coin from the first district, two from the second district, three from the third, etc. Then just look at the scale and see what number of nine appears after the decimal point.

If it's .9, it's the first district, if it's .8, it's the second, if it's .7 it's the third, if it's .6 it's the fourth, so on and so on.

I'm still working on the second one.

 
Comment by Martias on February 22, 2010 at 9:46 AM

I meant number, not "number of nine"

 
Comment by sophlightning305 on February 22, 2010 at 11:00 AM

=)

 
Comment by Martias on February 22, 2010 at 11:31 AM

Sorry, I was rushing out to get to class and I missed an obvious problem with my method.

To make things simpler, you should just see how much is missing from 78 ounces (the summation of all the coins). If it is .1, it's the first district, .2, it's the second, and so on. For my original method you would run into problems if district 11 or 12 were the district cheating you, because the number after the decimal point would be .9 or .8, which, if you looked at my method, would be the first or second district. With this revised method, you would be able to tell the difference quite easily.

 
Comment by Martias on February 22, 2010 at 5:00 PM

Ok, for the second one, I think I have an answer, though it seems kind of bogus.

The number on the blue house was 13. The children's ages are 6, 6, and 1. The reason why the man needed an additional clue is because the two ages are the same. If two of the children were fraternal twins, then they would have the same age, even though the one with the blue eyes was older.

This seems a little too much like cheating however.

 
Comment by sophlightning305 on February 22, 2010 at 6:45 PM

keep going

 
Comment by Kevin, NeuEve Team on February 22, 2010 at 7:39 PM

Martin, that's not correct. 6, 6, and 1 is not the answer because if two of the kids were both 6, then they would be exactly the same age, and there would be no "oldest" child. However since he says that his "oldest" child has blue eyes, there must be one, and only one, oldest child.

Think about it, what's the number on the house? If the house number is 13, then there are 2 possible configurations: 6,6,1 and 9,2,2, so the answer is ambiguous. That's why Guy #1 says "yo that's not enough info." That's why Guy #2 has to make it clear that there is only one oldest kid with that blue eyes statement.

Therefore the answer is 13, and the ages are 9, 2 and 2.

 
Comment by eohcnrk on February 22, 2010 at 7:56 PM

no no, you guys are both wrong, the ages are 3,3,2,2. the first dude had an illegitimate child...i think that's why this riddle was so hard...

 
Comment by Martias on February 22, 2010 at 11:00 PM

Oh yah...you are definitely right, K Tao. I missed that configuration =(

 
Comment by eohcnrk on February 23, 2010 at 12:28 AM

why is the sum necessarily 13? 3,3,4 would work as well, sum is 10 and you have one oldest child...or was ktao pointing out that sum does not need to be 13

 
Comment by a. kim on May 12, 2010 at 7:38 PM

just wanted to point out... u said the difficulty level was difficult haha... just sounded funny

 

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