The Mathematician and the Pea
You have 10 cans of peas. The cans are open. In each can, there are 100 peas. In nine of the 10 cans, each pea weighs one gram. In the tenth can, each pea weighs only 0.9 grams. You do not know which can has the smaller peas, nor is it possible to tell with the naked eye. To help you, you have an electronic scale. However, the scale is not in great shape, and can only provide one correct measurement before permanently malfunctioning. How can one, using only the one measurement afforded by the scale, determine beyond a shadow of a doubt which can it is that has the smaller peas?
Submitted by Chris Timmel.
To: Juliana Donatello: unfortunately your email address didn’t work. you are correct and can email me at sjm1 AT williams.edu.
email me at sjm1 AT williams.edu
i dont understand!
The reason is that the riddle says the scale is in bad shape and can only be used ONCE before breaking. sjm1 AT williams.edu
Can you please email me the answer? and why can’t I just add the first can, then the second and so on till I find the weight to be less than the number of cans x 100 gms? I mean, add the first can, if it’s 100 gm, add the next one and keep doing that until you find the lighter can.
Anonymous1: correct! sjm1 AT williams.edu
Derek: well done – email me at sjm1 AT williams.edu to chat
that’s a very general question: can y ou narrow it a bit? //s
hello steven..!!! i just wanted to ask you if you had any ideas on mathematical modelling??? the topic is mathematics in daily life…
yes, but that won’t work. //s
can you take out one pea from each box and weigh them together
Truman P: correct!
anonymous: correct, well done; not posting as it’s the soln
Sent a hint — if you want to chat further, email me at [email protected]
What is the answer?>
Hmm, please let me know if you are ever building planes or bridges so I can adjust my travel accordingly! There is a non-destructive soln….
After your free reading, break the electronic balance and use its parts as a ledge to weigh the remaining 8 cans – how many ever million times you want. 😀
(If you can’t think of a solution, make the question work for you! – I’m an eng. student 😛 )
P.S., thanks for all the mails you send me telling me if my answers are right/wrong. 🙂
Kareem: correct! not posting as it’s the soln
Jim: correct. My dream is to have the option to have people score the riddles and have that adjust the ranking.
No — any scale would work
Does it matter that the scale is electronic?
Aldo: correct. If you leave your email address, I can email you if you are right or wrong.
Whoops, misread what you wrote. You can only use the balance once for free. After that, each addition weighing costs $1million, cash up front.
Taking 1 pea from each can and putting them to electronic balance .
1 pea from first can then 2nd, third… And putting them on balance 1 after 1 in the order in which cans are placed. When the balance wil start showing 0.1 gram deficiency then the pea which was added just then to the balance will be the lighter one and the can to which it belongs wil b the required can.
ANT: correct
Mark: your answer was correct (not posting it as it gives the solution). Well done.
I’ll email you the hint, but here is one. You can’t add the cans one at a time. You need to take a special number of peas from each can. You take a different number from each can. Remember, if one of the cans is light then ALL of the peas you take from that can are light. Try to do a simple case when there are just two cans. If you can do this, try to do 3 cans….
I have the same question as k nolte.
I need a hint. can you add the cans one at a time while logging the weight increase?