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Posted

FishTeacher's thread on palindromes gave me an idea for some new number puzzles. Can you extend these sequences?

 

3, 5, 7, 9, 15, 17,.....

4, 8, 10, 13, 16, 20,....

5, 10, 15, 17, 21, 25,....

6, 12, 18, 24, 26, 31,.....

 

and a sort of related sequence the other sequences lead too,

 

1, 5, 27, 194,....

 

Now I think I have said too much already ;)

Posted

The first series are base 2 palindromes the second base 3 , ... . The last is special 1 is 1 in base 2 , 5 is 12 in base 3 , 27 is 123 in base 4, 194 is 1234 in base 5, ...

We could transform this into a game.

 

Somebody askes a question the first person who answers correct will ask the next question. If you see that in the last 24 hours there wasn't any post you will ask the next question, if there was one ( even a false response ) the first poster will answer the question and the author should give the solution to his question.

PS The questions should be math or physics related and not google questions.

 

 

Here is my question:

 

You have 12 coins. One of them has a different mass ( and you don't know if it is smaller or greater than the rest). Useing a balance and 3 measurements you must determine the coin and if it's lighter or heavier than the other 11 coins. This is a math question so please don't give a physics solution . ;)

Posted
The first series are base 2 palindromes the second base 3 , ... . The last is special 1 is 1 in base 2 , 5 is 12 in base 3 , 27 is 123 in base 4, 194 is 1234 in base 5, ...

We could transform this into a game.

 

Somebody askes a question the first person who answers correct will ask the next question. If you see that in the last 24 hours there wasn't any post you will ask the next question, if there was one ( even a false response ) the first poster will answer the question and the author should give the solution to his question.

PS The questions should be math or physics related and not google questions.

 

 

Here is my question:

 

You have 12 coins. One of them has a different mass ( and you don't know if it is smaller or greater than the rest). Useing a balance and 3 measurements you must determine the coin and if it's lighter or heavier than the other 11 coins. This is a math question so please don't give a physics solution . ;)

Of course, with the answers to the solution hidden others may continue working on the original problem. Although they now have the next number in the sequence they still have the puzzle of finding the next one after yours ;)

 

I don't know that it would be fair for me to answer the 12 coin problem since it is quite an old problem I've faced before. Perhaps I'll wait aq day or two and post a hint.

 

Speaking of hints, there is a hint to my sequences in every post I make ;)

Posted

Thanks for the edit. I didn't have time to look through the source of the page to find the color so I've selected the nearest color in the list.

 

Another rule. When you post your solution , post it like this

*color=#F6F8FA] solution */color] ( use [ instead of * ) , so that other users can solve the problems if they want without unwillingly seeing the solution.

Posted

Hum, there doesn't seem to be any guesses to either problem. Maybe it would help to begin each sequence as 11...

Start the 12 coin problem by weighing coins 1,2,3,4 against 5,6,7,8.

Posted
Split and weigh....6 vs. 6

Take the light stack, 3 vs. 3

Take the light stack and weigh 1vs 1.

If on is lighter it is false, if they balance the one left out is fake.

But you don't know if the counterfeit coin is heavier or lighter, only that it is different.

Posted
Nice, I like the hints. I've recently seen this problem too though, so it would be unfair for me to answer (It was a bonus question on a recent physics test).
Divide the coins into 3 groups of 4 and number them 1 - 12, not groups of 6

So, contribute a hint then :xx:

Posted

or you could name them after the way they influence the balance if they go up name them U coins , in they go down D coins, coins that you know they are ok G coins and suspect coins S. First there are 12 S and then ...

Posted
Divide the coins into 3 groups of 4 and number them 1 - 12, not groups of 6

So, contribute a hint then :xx:

 

Uhh, okay.

After dividing them into groups of three weigh group one against group two, there are now two possibilities: 1) They weigh the same and the "right" coin is in the 9,10,11,12 group and problem solved from there on out easily enough) or 2) They don't balance and then...

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