A Penny for Your Thoughts
You have 12 coins, one weighs slightly less or slightly more than the others. Using an equal arm balance and only making three weighings determine which one is different and whether it is slightly less or slightly more. Warning: the fact that you don’t know whether it is less or more is a major problem in the
solution.
Communicated by J. Miller.
your email address is not working — your post is wrong, email me at [email protected]
Doing 3 weighings only seems impossible. But this is how I’ll find that coin if I’d try my way to solve it
1. Weigh 6 coins on each arm of scale.
Since 1 coin is either lesser or more heavier compared to one of the other 11 coins with equal weights, then the weighing scale’s arms must be unequal in level after putting 6 coins on each side.
Assuming the heavier group of six coins contains coin X, then
2. we repeat the process, placing 3 coins on each arm and again taking the group that is heavier.
Only 3 coins remain. We took the heaviest group in every weighing we did.
3. Among these 3 coins, choose 2 and put one on each arm of the scale and see if they have the same weight. If yes, replace one coin with the other remaing coin of the last 3 coins. If the coin you placed last causes the scale to change in level, then that is coin x. If it does not caused any changes, we take the other 6 coins we first disregarded and do the same step.
email me at [email protected] — email to you bounced
Pls tell answer
My email to you bounced — please email me at [email protected]
what was the answer from your riddle? please I need it for educational purposes. thank you
There is an interesting twist to this problem. The weighings have to be independent, i.e., you have to specify upfront what is going to be weighed in each weighing and then, solely based on the results of the weighings, determine which penny is different from the others and whether it is lighter or heavier than the rest.
I’ll bump up the difficulty. My dream was to have the site set up so people could vote and adjust the level accordingly.
This is not a medium difficulty riddle. I remember when i first encountered it, some years ago, took me days to figure it out. Its definitely hard, at least.
sadly no
you can’t put 4 on a side and remember the weight as you can only weigh
with items on both sides. //s
put 4 on one side of scale and remember weight. put four on other side. there the same so discard those 8. put 2 on same side of scale and see if that weight is half of the first 4.
if so discard those 2. put 1 on one side of scale, if it’s weight is half of 2 youve found your coin
sure, glad you’re enjoying
I have just found this website. The kids are loving these. Can you please send the solution for finding whether its heavy or light
Your style is unique compared to other folks I have read stuff from. Thanks for posting when you’ve got the opportunity, Guess I will just bookmark this blog.
Hint sent (sjm1 AT williams.edu)
Can you please send me your solution?
I don’t believe this works as we don’t know whether or not the coin is heavy or light. Email me at sjm1 AT williams.edu to chat more. ..s
Weigh 3 coins against 3 coins. If they balance, they are all good coins, if they don’t, the ones not being weighed are all good coins. Now there are 6 good coins and a group of 6 containing one bad coin. Weigh 3 good coins against 3 potential bad coins. If they balance, the remaining 3 contain the bad coin. If they don’t balance, the bad coin is in that group of 3 and it will be obvious if it’s lighter or heavier. Now there are 9 good coins. From the remainig 3 coins, take 2 and weigh them against each other. If they balance, the third one is the bad one. If they don’t, the bad coin is the one that is lighter or heavier (depending on the results of the second weighing).
I’ll send a hint first. //s
Can I have the answer,plz?
if you want I can send my soln. you have to get used to breaking things into cases where you know more and can use that info to attack the problem.
just email me at [email protected]
I don’t know if I just suck at those coin-weighing riddles or if this one is above ‘medium’, but it was a really tough nut to crack. I did it, eventually, but not until I’ve spent almost two hours getting my brain evaporated through my ears. This really is a challenging one since the strategy is different depending on the result of the first weighing.
I’m having trouble understanding all that you wrote. YOu are welcome to read the riddles and post comments on the pages, or email me directly at this account: sjm1 AT williams.edu. //s
Magnificent beat ! I wish to apprentice at the same time as you amend your web site, how could i subscribe for a blog web site? The account helped me a appropriate deal. I had been a little bit familiar of this your broadcast offered brilliant transparent concept
ok (sjm1 AT williams.edu)
please send soln. I give up lol
sure //s
Would you mind sending me the solution to this penny problem?
Thanks
nope — email me at sjm1 AT williams.edu if you want to chat more about this. //s//
The answer is pi.
the problem is you don’t know if the different coin is heavier or lighter — your soln works if you know the odd coin is heavier.
in the first attempt we have 6 on each side. the one heavier would have the heavy coin. Now from the heavier batch of coins we select four coins and weigh with 2 on each side. if they are equal then on the last attempt we check the other 2 coins the one heavier would be our coin. If they aren’t then we would way, then we would take the set of two coins which is heavier and weigh with each coin in each side of the arm balance. the heavier coin would ours.
no — start with 4 and 4….
actually you’ll get this problem if you got a balaned scale (not unbalanced scale) .. what if you remember when testing 3 on 3 from the bad pile, which one weigh less (or more) … now if you choose the same set of 3 that you have tested in step one you will be able to decide if the coin weighs less or more .. I’am right ?
hint: start with 4 and 4
use the fact that after teh first weighing there are coins of known correct weight
Problem is that we don’t know the odd coin out is weighing less or more as compared to other 11 coins. How do you say that the one which weighs less is the penny we are looking for…??
not positive, but don’t think it can be done 6-6; I started 4-4. //s
This is what i have been thinking so far…. am I on the right lines?:
Start by halving the coins into two 6s. Choose one of these sixes and divide into half and place 3 coins on one side, and 3 coins on the other. If the coins balance equally, then these coins can be disguarded, or if they dont, then keep them to re-weigh???
sure, will send //s
We divide by 4 groups of 3 (1,2,3,4)
We weigh 1 and 2 — Weigh No. 1
– If 1 is not equal to 2 then we weigh 1 and 3 — Weigh No.2
If they are equal then the coin is in group 2 and we know if 2 is lighter or heavier than other groups
From group 2 we take two coins out of it and weigh them — Weigh No. 3 — if they equal that the coin which has been left out is the coin we are looking for if not than one of the two weighed coins will be the different one (we know already if it is lighter or heavier)
If 1 and 3 are different the the coin is in group 1 and we know if it heavier or lighter depending on the result of the previous weigh and we do the same exercise for the three coins remaining
– If 1 = 2 then we weigh 3 and 1 — Weigh No. 2
If 1 is not equal to 3 then we know that 3 contains the coin and we know if it is lighter or heavier
We take group 3 and we take two coins out of it and weigh them Weigh No. 3 if they equal that the coin which has been left out is the coin we are looking for if not than one of the two weighed coins will be the different one (if 3 is lighter than 1 then we take the lighter coin and vice versa)
The problem is if 1 is equal to 2 and is equal to 3 then we cannot know if the coin in the group 4 is lighter or heavier.
Can you please share with me the solution by email?
Thx
Terrific — feel free to also email me at sjm1 AT williams.edu //s
Steven,
Thank you for the response. Now you will see me here more often.
Sure. //s
Hi,
Excellent web site, i had been looking for something like this for a long time.
Could you send me the answer to this one, please?
sure ..s
Hi my friends and I have been trying this problem but so far have been unable to figure out a solution could you please send me the answer? Thanx
sure, will send ..s
try 4 and 4….
Ive been trying to solve this can I have the answer
My possible solution is to divide the 12 by 2 and place 6 coins on each side, then the heaviest redivide by 2 until you place 1 coin on each side of the balance and holding one separate. If the one is a different weight it would be indicated but if it balances out the one that is not placed on the balance is the odd weighted one. That is when I got stuck…
I’ll email you my soln and you can compare, but they’re probably the same. //s
That is really a classic one. I just emailed the solution to you. Think that is a correct way of doing it. Please let me know if you have any easier method.
more detail is needed as there are complicated subcases — email me at [email protected] if you want to see my soln. /s
separate into 3 groups of 4. measure 2 of the groups. if they are equal then it must be in the third group. separate into 2 groups of 2. find out which group is heavier then weigh the 2 coins.
done — email me at [email protected] if it didn’t arrive
mind sending me the answer to the riddle?
I dont’ think this works — in some situations I’m not sure if you can correctly identify if the coin is heavy or light.
Step one: Weigh two groups of 5 coins and leave 2 coins off to the side:
Step two: Take 4 coins from the lighter/heavier side from from Step One and place 1 coint off the side, weighing only 2 groups of two.
Step three: Finally, weigh only 1 coin on each side using the two lighter/heavier coints from Step two.
If when doing Step 1 the weighs would have been equal, you could have known it was between the two you originally left to the side.
If when doing step two the weights were equal, you could have weighed the side pair from step one to find it, or if those were equal simply inferred that the single coin left to the side AFTER step one was the special coin.
After doing step 1 and having the weights be off, you would have known that neither of the two originally set-asside coins were NOT the special coin.
This should work.
Kareem: your solution looks correct. If you email me at [email protected] I’ll send along mine.
sure — email me at [email protected] to submit
I have a math riddle that I want to submit to this website. Can I submit it? and if I can, how?
This is one of the harder ones in the category. Email me at [email protected] if you want a hint.
I actually think this riddle is harder than most of the ones in the hard section of this website.
sure — I’ll send you an email
Can I have the answer?
sure, emailing it to you.
Can I see the answer to this one?
Right, unfortunately there’s that problem. I can do it by starting 4 on 4.
1st Step:
– Take 6 coins at random from the 12
– Split them into two groups of three and weigh across the balance.
If they are in balance, set aside these 6 coins as the “good” pile (and the other 6 as the “bad” pile). If they are not in balance, then set the random 6 as the “bad” pile.
2nd step:
– Take 3 coins at random from the “bad” pile and 3 from the “good” pile and weigh them across the scale.
– If the “good” and “bad” piles are not in balance, then the randomly selected 3 from the “bad” has a defective coin. This will also tell you whether the defective coin is less or more. If in balance, then the 3 not selected at random contain the defective coin.
3rd step:
– From the defective 3, take 2 coins at random and weigh across the balance.
– The defective coin will weigh more or less (as judged by step 2). If in balance, then the 3rd non-selected coin is the defective.
However, I’ve now realized that unless you’re able to conclusively prove whether the defective coin weighs less or more in step 2 (by getting an unbalanced scale), you won’t be able to decide conclusively on the defective coin in step 3.
Interesting. I didn’t think 6 of 2 work — if you have a soln and it’s simple to write up, can you send it to sjm1 AT williams.edu?
Might I just say, instead of 3 groups of 4. 2 groups of 6 is alright to use, just a hint here…
little worried — I did it with a different grouping
4groups of 3.
1 2 3 4 5 6 7 8 9 10 11 12
First measure 1~6. 123 in one side and 456 in the other.
If You see a difference, for example if the (123) is heavier and (456) is lighter,
U know there is a heavier one or lighter one in one of both sides.
Therefore you take (157) and put it in one side and (428) in one side.
Now you have 3 different numbers from different groups.
Numbers 1 and 2 comes from the group that weighed heavier
Numbers 5 and 4 comes from the group that weighed less
Finally numbers 7 and 8 comes from the group that has the normal coins.
Now if 157 still weights more than 428, U know that 1 is heavier or 4 is lighter.
Therefore now you take 1 or 4 and compare it with 7~12 (any coin is alright).
Now you know the answer.
If 157 weights less than 428, u know that either 5 or 2 is the answer.
Therefore you compare it with 7~12 (any is fine).
If 123 and 456 weights the same
You now measure 789 101112.
If 789 weighs more than 101112
You put 7,10,11 in one side and 8,1,2 in the other side.
( You notice we left out 9 and 12).
If (71011) still weights more and (812) weights less,
then we know that 7 weights more.
If (71011) weights less and (812) weights more,
Then we know that 10 or 11 weights less or 8 weights more.
Therefore you just measure 10 and 11. if they’re equal u know that 8 weights more.
If they’re not, well u know that one of them weights lesser.
After some weighings, though, you will know a few coins that must have the correct weight.
I had thought of that Richard but if you don’t know if the penny weighs more or less then how could you know which group it’s in?
You’re on the right path to the solution.
No problem — feel free to email me solutions at [email protected]. We’re still trying to figure out exactly what we want to do on the webpage. For now, it’s more general discussion about the riddles (such as the moat should be rectangular for one of them), or how these might be used in a class.
sorry i didn’t know that i was not suppose to leave the solutions in the comment area
thanks for letting me know i got the answers right i wont post the answers anymore
Richard C: I don’t want to post your response as it contains the solution. If you email me at sjm1 AT williams.edu I can comment on your solution and share mine as well.
you divide the coins into 3 groups of 4 take two of the groups and measure them then if they’re equal then the coin must be in the third group if one weighs less the coin must be there. then now knowing which group it is separate the pennies into groups of two one should weigh less take that group and measure the last two pennies and the one that weights less is the penny you were looking for