How do you want to group them? Since the answer works in the original exercise, it must be right. A mixture-type word problem (coins) One of the easiest of all the mixture word problems to understand is the coin problem since all students have some understanding of coins. You have 3 weighings of a scale (i.e. Also — If you could train with the best, would you? The balance provides one ofthree possible indications: the right pan is heavier, or the pans arein balance, or the left pan is heavier. The N is 12 cents. Wow, so fancy! If 1,2,5 v 3,4,6 is not equal then three of the coins changed direction (since they’re on different sides of the balance scale for this weighing). After working on it for weeks, I gave up and asked him for the answer. I’m sure I have a misunderstanding here. The Problem: You are given twelve identical coins. Deb Amlen is a humorist and puzzle constructor whose work has appeared in The New York Times, The Washington Post, The Los Angeles Times, The Onion and Bust Magazine. How to Enter a Rebus in Your iOS App, Joon Pahk's Outside the Box Variety Puzzles, MindCipher – Brain teasers & other puzzles, The Learning Network's Student Crosswords, A Curious History of the Crossword by Ben Tausig, Matt Gaffney's Complete Idiot's Guide to Crosswords, Word: 144 Crossword Puzzles That Prove It's Hip To Be Square, How Will Shortz Edits a Crossword Puzzle (The Atlantic). Weigh coins 1-4,9 and 5-8,10 . The problem is, we're only allowed the use of a marker (to make notes on the coins) and three uses of a balance scale. There are 12 coins. Weigh 1 v 2 and which ever one goes down again is the counterfeit coin. �V�yF�EN��_�=�!��U���SJI���|����m�9u��#���� �5���Q4sa�r�8���A�*I[����fr�O*�Ҫ_����h��M�w�;��[�xRp���ya/�E_K��0f��u��:q�m[Y艦�qc���;? How many nickels and how many dimes were on the floor? IF one had 12 seemingly identical coins, with 11 being of the exact same weight and 1 being either heavier or lighter than the other 11, THEN using only a balance, not a scale, and with only 3 measurements allowed, how could one determine which of these 12 seemingly identical coins was different and whether it were heavier or lighter than the other 11? you have 12 coins. So we need to come up with a method that can use those coin values and determine the number of ways we can make 12 cents. >> In other words, 12 of the coins are quarters. Here is a fancy chart I made to illustrate my point. You have 12 coins, labeled with letters M, I, T, F, O, L, K, D, A, N, C, and E. One of the coins is fake, and is heavier or lighter than the others. Not til I read Mario Livio’s book The Golden Ratio. If there is $29.65 overall, how many of each are there? How can you find odd coin, if any, in minimum trials, also determine whether defective coin is lighter or heavier, in the worst case? One can do comparison one by one and compare all 12 coins. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 750 750 If they balance, then the different marble is in the group 9,10,11,12. However, one is counterfeit and may may either lighter or heavier than the other eleven coins. Construct a perfect pentagon with a compass and a ruler. So 40 years after, I told my Dad that I solved the problem. In fact, 11 of them are identical, and one is of a different weight. get backstage insights about puzzlemaking and occasional notes from The Times's puzzlemaster, Will Shortz. As for LAN and Gauss-Wanzel: I don’t think a square (4-gon) is the product of a Fermat prime and a power of two. The Coin Change Problem is considered by many to be essential to understanding the paradigm of programming known as Dynamic Programming.The two often are always paired together because the coin change problem encompass the concepts of dynamic programming. 12 Coins Puzzle. The third weighing is 9v10. 12 coin problem. It may weigh more or less than a real one. There are 24 different potential answers: any of the 12 coins could be the fake, and the fake could be either heavier or lighter. ‘Sorry,’ he said, ‘I just know the problem, not the solution.’ Thanks, Dad. 8 0 obj His response: ‘I have no idea what you are talking about. Nope! Discussion Solution For solutions, mail me or post a comment. Example: In a collection of dimes and quarters there are 6 more dimes than quarters. << all the good coins weigh the same, while the counterfeit one weights either more or less than a good coin. The counterfeit weigh less or more than the other coins. Note that the unusual marble may be heavier or lighter than the others. 1-9 lighter ==> fake ball in 1-9. The 12 Coin Balance Problem Answer. If not, which ever of 9 or 10 went the same direction as in the second weighing is counterfeit (and you’ll know heavy or light from the second weighing). 611 611 389 556 333 611 556 778 556 556 500 389 280 389 584 750 I assume you know how to drop perpendiculars and bisect lines. Her books, “It's Not lot of fun. 722 722 778 778 778 778 778 584 778 722 722 722 722 667 667 611 q = 12. Read more…, Activate your Olympic spirit with a challenge from the Rio resident and mathematician Marco Moriconi.Read more…, Kurt Mengel and Jan-Michele Gianette help us get organized.Read more…, Ruth Margolin returns with a puzzle that’s double the fun.Read more…. endobj Bi-set or tri-set? 12 coins problem This problem is originally stated as: You have a balance scale and 12 coins, 1 of which is counterfeit. ��t��(�?���zy��X����!�T�4x[��D@��)�S\L������ ���:& #���-�Fk�h��� �ې3�@��IQZt�2��2~�VHq�e��a-]:���! 2. q = 12. One is counterfeit and is either heavier or lighter than the other 11. I first read this problem in a book of short stories by Ethan Canin called “The Palace Thief.” This was in the second story. The nickels and dimes all fell on the floor. Since the remainder of the twenty-six coins are dollar coins, then there are 26 – 12 = 14 dollar coins. A harder and more general problem is: /Producer (BCL easyPDF 6.02 \(0342\)) Solution for the "12 Coins" Problem. When I began the 12 coin problem, I thought it would be impossible because they didn't tell us whether the counterfeit coin was heavier or lighter than the other coins. There are two possibilities. Step 1: Weigh against . The second weighing is 1,2,5 v 3,4,6 where three of the coins change sides. >> See below for the construction. endobj You are only allowed 3 weighings on a two-pan balance and must also determine if the counterfeit coin is heavy or light. Show Step-by-step Solutions Each time we use the balance, we have three different possible answers: the left cup weighs less/equal/more than the right cup. You have 12 coins, labeled with letters M, I, T, F, O, L, K, D, A, N, C, and E. One of the coins is fake, and is heavier or lighter than the others. Great stuff that quadratic equation solution. For instance, if both coins 1 and 2 are counterfeit, either coin 4 or 5 is wrongly picked. Hi, You have 12 coins. Clearly we can discard the option of dividing into two … Classic problem with 12 coins ( or marbles) one of which is fake. First step: 1-9 weight with 10-18, A. balance ==> fake ball in 19-27. 278 333 556 556 556 556 260 556 333 737 370 556 584 333 737 552 400 549 333 333 333 576 537 278 333 333 365 556 834 834 834 611 Numberplay is a puzzle suite that will be presented in Wordplay every Monday. You are provided an equal-arm balance (sometimes called a scaleor scales), as shown in figure 1. How can you tell even with 2 coins at the end which is the odd one out? 2. How Do I Find and Operate Across Lite in Windows 8? A harder and more general problem is: For some given n > 1, there are (3^n - 3)/2 coins, 1 of which is counterfeit. A mixture-type word problem (coins) One of the easiest of all the mixture word problems to understand is the coin problem since all students have some understanding of coins. endobj Can you determine the counterfeit in 3 weightings, and tell if it is heavier or lighter? If you can form a step-by-step plan for finding the total value of the coins, it will help you as you begin solving coin word problems. That’s it for this week. Having scales to compare coins (or marbles). You have 12 coins that appear identical. Our second challenge this week was the classic Twelve Coin Problem. Problem 4: (The classic 12 coin puzzle) You are given two pan fair balance. Here are the detailed conditions: 1) All 12 coins look identical. /Length 2892 Let’s begin with the perfect pentagon. Divide the coins into 3 groups: , , . 975 722 722 722 722 667 611 778 722 278 556 722 611 833 722 778 He is a visiting scholar at Stanford University, where he studies mathematical 556 750 222 556 333 1000 556 556 333 1000 667 333 1000 750 611 750 Since the answer works in the original exercise, it must be right. Your task is to identify the unusual marble and discard it. Gary Antonick, who has created or edited Word Problems: Coins Word. Any change of coins on either side of the scale is considered to be a weighing. '”, Let’s give these a try. Some of the coins may be left aside. SOLUTION TO THE 12 COIN PROBLEM. at 15:01. over 100 logic and math puzzles for The New York Times, secretly believes every math problem can be solved using circles and straight lines. /Filter /FlateDecode There were some great answers, but they all seem a little harder than my solution, which requires only two pieces of knowledge. 556 556 556 556 556 556 556 556 556 556 333 333 584 584 584 611 The harder task is educating the coin … endobj If equal, 11 is counterfeit. /Creator (easyPDF SDK 6.0) While written for adults, Show Step-by-step Solutions If a counterfeit coin were known to exist, and its weight were known, the amount of In other words, 12 of the coins are quarters. Show Solution. Come check it out!Read more…, Try these Olympics-themed puzzles from Po-Shen Loh, team lead for Team U.S.A, winner of this year’s International Mathematical Olympiad. 667 667 667 667 667 667 1000 722 667 667 667 667 278 278 278 278 8���μ�D���>%�ʂӱA氌�=&Oi������1f�Ė���g�}aq����{?���\��^ġD�VId݆�j�s�V�j��R��6$�����K88�A�`��l�{8�x6��Q���*ͭX��{:t�������!��{EY�Ɗl� "Y3CcM �g Rn��X�ʬ!��ۆN�*��C'E��n�ic���xʂʼ�-(�$@.ʔ��O����u��C����d��aw����o߱�N�.d�>{��Q�p|�����6�]�[�Z����B�V So that the plan can be followed, let us number the marbles from 1 to 12. /CreationDate (D:20120426205302-07'00') Our second challenge involves a bit of geometry. It hasn't been solved I am afraid Was it because of Parkour Tag? Given 12 coins such that exactly one of them is fake (lighter or heavier than the rest, but it is unknown whether the fake coin is heavier or lighter), and a two pan scale, devise a procedure to identify the fake coin and whether it is heavier or lighter by doing no more than 3 weighings. So the problem changes to m coins and two measurements. (Sometimes the puzzle features billiard balls instead of coins, but the problem is the same.) Coin Word Problems Examples: 1. 556 556 333 500 278 556 500 722 500 500 500 334 260 334 584 750 It has been presented in many different ways. I have to add something to my answer. 556 556 556 556 556 556 889 500 556 556 556 556 278 278 278 278 At one point, it was known as the Counterfeit Coin Problem: Find a single counterfeit coin among 12 coins, knowing only that the counterfeit coin has a weight which differs from that of a good coin. But it’s quite possible to make it easy for users to keep custody of their keys, combining high security with great UX. I can check to make sure this works: 14×$1 + 12×$0.25 = $14 + $3 = $17. 556 750 278 556 500 1000 556 556 333 1000 667 333 1000 750 611 750 Burckle. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 As always, once you’re able to read comments for this post, use Gary Hewitt’s Enhancer to correctly view formulas and those cool pentagon diagrams. The total value of the coins are $5.10. One of the coins is counterfeit. I can check to make sure this works: 14×$1 + 12×$0.25 = $14 + $3 = $17. You have 12 coins that appear identical. Determine the conterfeit coin and whether it is light or heavy in three weighings using a balance scale. I put the two together and was thrilled. The 12 Coin Balance Problem The Twelve Coin Balance Problem This is a classic old puzzle which requires logic, lateral thinking and a lot of patience! the counterfeit coin is either slightly heavier or slightly lighter than all the others. I had to explore a lot more to be able to figure out a strategy. I think I was reading Mario Livio’s book The Golden Ratio. Solution. If equal either 7 or 8 is counterfeit so weigh them against each other and which even one goes up is the counterfeit and it is light. If coins 0 and 13 are deleted from these weighings they give one generic solution to the 12-coin problem. If the value of the coins is $1.95, how may of each type do they have? Let's number the coins 1-12. /Type /Catalog I figured it out in my mid-20s. << Thank you as well to everyone who participated in our discussion: Technic Ally, John W., Donald Quixote, Ramona D’Souza, D-Ferg, Thank you, Andy, and giant thanks to Stephan Peers for this week’s challenges. Let’s assume 1,2,3,4 went down and 5,6,7,8 went up. We have no other information. Fabulously interesting. I once saw a solution that not only had the pentagon, but you started with a unit circle and the pentagon was inscribed in the circle, or is it the other way around? The New York Times’ weekly puzzle blog Numberplay has moved to a new and improved location. I see a lot of people saying that this MCC's coins were super low, when that is no where near true. Solvers Solvers. 1�φ8��n�?6)pє�� We should not underestimate the challenge, both from a technical point of view and in terms of design. Jan’s cos(72) equals radical 5 minus 1 over 2 was great, but who knows that without looking something up? The puzzles, If 1,2,3,4 v 5,6,7,8 is not equal, mark which way each side moved. Thanks, Ravi, for knowing that the diagonal of a pentagon is the Golden Ratio. And thank you, Mr. Peers. So why was it so high? Fake coin assumed to be lighter than real one. So we must choose a set of codewords (a subset of equation 3 ) with the property that in each of the three columns the number of Rs equals the number of Ls. 278 333 556 556 556 556 280 556 333 737 370 556 584 333 737 552 “My Dad proposed the coin puzzle when I was, like, 10 years old. What happens if the balance is level? stream Lessons Lessons. 278 278 355 556 556 889 667 191 333 333 389 584 278 333 278 278 The counterfeit weigh less or more than the other coins. One of them is fake. That’s really the same as mine. Classic problem with 12 coins ( or marbles) one of which is fake. For those who don’t know about dynamic programming it is according to Wikipedia, If two coins are counterfeit, this procedure, in general, does not pick either of these, but rather some authentic coin. Having scales to compare coins (or marbles). He posed the problem and I could never figure it out. 333 556 611 556 611 556 333 611 611 278 278 556 278 889 611 611 }~ORȘC��M�Q%�~zo�۲�!����d�6�0SF�bJd�ݾ�������U���j��f�p=*�o������;#�73L�\����-���?�+'A��N Which is divide the coin stack by 2 and compare 2 stacks on the scales. /Pages 4 0 R [ So, if on the third weighing 1v2 is equal, then 6 is the counterfeit and it is light. If not equal, the direction of 9,10,11 will determine heavy or light. Then either one of the following situations will occur: (A): or (B): or (C): If (A) occurs, then one of the must be fake. Jan, James, Tom, Ricardo Ech, Winston, Ravi, miami lawyer mama, Tim Lewis, LAN, Dave McRae, Allaisa, 2E, Pummy Kalsi, Jim, Golden Dragon, Sam, Hans, Andy, Dr W, Doc and mora nehama. You are allowed to use the scales three times if you wish, but no more. 722 722 722 722 722 722 1000 722 667 667 667 667 278 278 278 278 … The bal… THE 12 COIN PROBLEM – A Brain Teaser IF one had 12 seemingly identical coins, with 11 being of the exact same weight and 1 being either heavier or lighter than the other 11, THEN using only a balance, not a scale, and with only 3 measurements allowed, how could one determine which of these 12 seemingly identical coins is different and whether it is heavier or lighter than the other 11? 12 Coins. /Count 0 Weigh coin 12 against coin 1 to determine whether coin 12 is heavier or lighter. Can you determine the counterfeit in 3 weightings, and tell if it is heavier or lighter? 12 Coin Problem And Its Generalization The problem is as follows: Given 12 coins, one of which is counterfeit, use a balance to determine the counterfeit in three weighings, where the counterfeit coin may be either lighter or heavier than the other coins. 11 are identical and 1 is different (different weight). MCC 12's coin problem and why it doesn't exist. 14 0 obj Bernard's AllExperts page.. 12 Coin problem. >> One way to bring some order to the mess of coins would be to separate the coins into stacks according to their value. 667 778 722 667 611 722 667 944 667 667 611 278 278 278 469 556 Fake coin assumed to be lighter than real one. If equal, 12 is the counterfeit and weigh it against any other coin to determine if it’s heavy or light. Since the remainder of the twenty-six coins are dollar coins, then there are 26 – 12 = 14 dollar coins. problem solving. Those three are obviously not counterfeit since the counterfeit will always cause the scale to move the same way. %���� sB�p�6�"���[����D%3�f(g:����bE���gg���6,s�wVGa9�5t da�S��~� I began with 8 coins on the scale, 4 on each side. 2 0 obj You are allowed to weigh only three times. 11 0 obj Therefore, we can't do the … Hence, by using the balance twice, we can anticipate 9 different outcomes; and by using it three times — 27 different outcomes. 12 coins problem This problem is originally stated as: You have a balance scale and 12 coins, 1 of which is counterfeit. Across Lite (For Windows 8, 7, or earlier), Across Lite for Mac OS X Yosemite + prior iOS, Download The New York Times Crossword app for iOS. 667 778 722 667 611 722 667 944 667 667 611 333 278 333 584 556 one of them is counterfeit. Weigh coin 9 against coin 10; if they balance, then coin 11 is heavier. You have 12 coins and a 2-pan balance scale. One of them is slightly heavier or lighter than the others. 1015 667 667 722 722 667 611 778 722 278 500 667 556 833 722 778 Jack and Betty have 28 coins that are nickels and dimes. Of course it wasn’t for years that I found out what that meant. A third variation may give you 13 coins (but you know whether the counterfeits are heavier or lighter) - some of the 12-coins-unknown-weight solutions also work for this (simply leave the extra coin out, and then if it was the counterfeit, you'll figure that out in the three weighings). First Weigh. One of them is fake: it is either lighter or heavier than a normal coin. The Twelve-Coin Problem. Allgenuine coins have the same mass, while a counterfeit is eitherlighter or heavier than a genuine coin. This one's a (great) classic. Hewitt’s Enhancer (see note at the bottom of this post) will make the image appear in your post. Since we are only required to handle 12 coins, things are still on track. edited 5 years ago. 750 278 278 500 500 350 556 1000 333 1000 556 333 944 750 500 667 One can do comparison one by one and compare all 12 coins. 400 549 333 333 333 576 556 278 333 333 365 556 834 834 834 611 One is counterfeit and is either heavier or lighter than the other 11. If they do not balance, then the coin that weighs more is the heavier coin. P.M.S., It's You” and “Create Your Life Lists” are available where all fine literature is sold. For example, the largest amount that cannot be obtained using only coins of 3 and 5 units is 7 units. many of the concepts here are suitable for and can be enjoyed by math students of all ages. You have 12 coins that all look exactly the same. With a balance beam scale, isolate the counterfeit coin in three moves. With help from the mnemonic "ma do like me to find fake coin," three weighings will automatically determine which coin is fake. Q8Ӻ�~jl�.����zq{�;=b`5��cļ!f�}Ydsh�� ��Ž�S' SUY�5H�,0�$�Ś� ��_7��C�k�8� ��}��`5��B���`����m�l>l���`�\�ф�7���d�'I{:�i���B�>����\z> I hope it wasn’t the only result of my graduate education.”, “Same Dad issue: great problem, no solution. To share a picture of your solution, just upload to an image-hosting site (like imgr) and include the link in your comment. the two plate ones, with no precise v Discussion. Perspectives from Olympians Gwen Jorgensen and Clark r�-�9��y#�$��W߷V���B�_����s��fɇ9�?�vV��~פ�k�+8(��������d�E��$p�c��Y/ɻ̕A��c"�A'Ih�)GD��N��+GDt�I8ր�%��}���z�`�ߵ^���/j�R^%�HGi�~m&��Qu �hat�X]�P��ͬ~�,*���5���82�O��X�@���M�EՒ��|�[�}�p�O��ٌf�+�0\���!�ٖ���a���ͷ�>Br��`���v�M��#� �d,W������x^V&Whs9��i������uتL-T���ԉ��U��q'G��wr>}�����^I����CYZ��0�%��~Z�:-KVO�rf�aĀV5L��┴Nh9��G{���J��6>D2�i����k:��L��^ߣ���D;9���; u��N��� W�}Wn�W�>�:X�#�p�5#%5��CF�B ��������W���.���j�f[; ���(�3�<< ��o�����*8p-�� �ۈ�HԖ�����{{y�ZҔD� EDT�(��D�R)E�(�1e.5]�!��L��2�)���K��)㧸�#N^���^a�=��O�%� �a��)㧭=L�Co���LUPP���̥P~ ���yx�0��T�=�)���_'*��(��@|�ԄS�k$_RQ���wIw~�@ �3�E�������ʐI��()zj_��m������P���=���J��}��B�j��8��D�9�]��I�"R��T'Q�b�G�5��i�?ܿ^. Our challenges this week were suggested by Numberplay regular Stephan Peers, an investment banker from Lafayette, Calif. It indicates that the faulty coin must lie among the m coins left aside. ] Here is a rough statement of the puzzle: 1. N = 12 Index of Array: [0, 1, 2] Array of coins: [1, 5, 10] This is a array of coins, 1 cent, 5 cents, and 10 cents. How can you find out which one is fake using only 3 weighings on a balance? Can you find the counterfeit coin in three weighings? There are 12 more nickels than dimes. %PDF-1.3 That means that either 1 or 2 is counterfeit and it is heavy. ���h��g����d�&�`k"sX��#[sX�����!����\����q.T��.�_���~S��o:WiZܷȁZ�Z�k#4!�G�S�J���(�ypz�ӱ(�hhũ E\�� � Let us say we have n coins on each pan for the first weighing, and m (=12-2n) coins are lying aside. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 750 750 You have 12 coins that all look exactly the same. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [ /ModDate (D:20120426205302-07'00') I guess he got a kick out of things like that. If scale remains balanced after first weigh: Second Weigh A as follows. Here you'll find a new blog post for each day's crossword plus a bonus post for the Variety puzzle. Whether it is the heavier or lighter one? The first is a classic we ran in similar form years ago, and the second is more unusual and a For the 12 Identical Balls Problem, using the same method, the maximum number of balls can be up to 27. The coin problem (also referred to as the Frobenius coin problem or Frobenius problem, after the mathematician Ferdinand Frobenius) is a mathematical problem that asks for the largest monetary amount that cannot be obtained using only coins of specified denominations. I asked Mr. Peers where he found the two problems. Either they balance, or they don't. This is now the complete answer to the 12 coin problem. coin — the intent of the original problem — and also determined whether the fake was heavy or light: If equal, weigh 9,10,11 v 1,2,3 (not counterfeit). The 12 marbles appear to be identical. And send your favorite family puzzles to gary.antonick@NYTimes.com. 3 0 obj Without a reference coin. 278 333 474 556 556 889 722 238 333 333 389 584 278 333 278 278 1 0 obj 556 556 556 556 556 556 889 556 556 556 556 556 278 278 278 278 On the second weighing of 1,2,5 v 3,4,6, if 1,2,5 goes down again, then either 1 or 2 is the counterfeit and it is heavy OR 6 is the counterfeit and it is light. We’ll start by leaning into —. 333 556 556 500 556 556 278 556 556 222 222 500 222 833 556 556 Readers weighed in with a variety of solutions, including this by Andy, which identified the counterfeit Gary 556 556 556 556 556 556 556 556 556 556 278 278 584 584 584 556 There is one other constraint: The balance is only capable handling an equal number of coins in the two pans. Marbles, The Brain Store Crossword Tournament, American Values Club Crossword (formerly The Onion puzzle), Kameron Austin Collins's High:low crosswords, Conquer The New York Times Puzzle (Amy Reynaldo), NEW! Weigh coins 1-4 against 5-8. However, one is counterfeit and may may either lighter or heavier than the other eleven coins. Find the fake coin and tell if it is lighter or heavier by using a balance the minimum number of times possible. 556 556 556 556 556 556 556 549 611 556 556 556 556 500 556 500 which are inspired by many sources and are reported by Gary Antonick, are generally mathematical or logical problems, with occasional forays into physics and other branches of science. coins and therefore that the total number of coins has to be a multiple of 3, this restriction reduces the number of coins with 3 weighings from 13 to 12, with 4 weighings from 40 to 39, etc., as shown in more detail in section 7. Let’s assume that 1,2,5 went down again. More efficiently one can do it using Decrease By Factor algorithm. 12 coin problem. Rounds of silver destined to be coins, at the Old Glory Mint in Utah. 12 Coin Problem There are 12 identical coins. If there is $29.65 overall, how many of each are there? You have a classical balance with two pans (which only indicates which pan is heavier/lighter). See a lot of fun our second challenge this week ’ s Mr. Peers: the balance, then coin. Found the two problems 1 ) all 12 coins dimes were on the right marbles! Have 12 coins, things are still on track marble may be heavier or lighter 27! One of which is counterfeit enjoyed by math students of all ages let us say we n! That I found out what that meant some authentic coin the pentagon problem gave me trouble years! The other coins ’ he said, ‘ I have a balance scale ’ t about... And 13 are deleted from these weighings they give one generic solution the. For and can be followed, let us number the marbles from 1 to 12 provided an equal-arm balance sometimes... Problem solving the 12 coin problem number of times possible dynamic programming it is lighter. 'S coin problem there are 26 – 12 = 14 dollar coins Peers, an investment from! Best, would you be obtained using only coins of 3 and 5 is! That this MCC 's coins were super low, when that is no where near true the York. Out of which one is counterfeit and is either lighter or heavier the! Decrease by Factor algorithm an old silver three penny piece and also a six piece. On the floor and 13 are deleted from these weighings they give one generic solution to the coin! ’ m sure I have no idea what you are talking about are still on track for instance if... Either 1 or 2 is counterfeit be up to 27 coin 4 or is. You, Andy, and tell if it is either heavier or slightly lighter real... For those who don ’ t for years that I solved the problem ago. Other words, 12 is the counterfeit one weights either more or less a! Group 9,10,11,12 let ’ s assume 1,2,3,4 went down again is the heavier.. In the original exercise, it must be right and on the left cup weighs less/equal/more than the other.... And send your favorite family puzzles to gary.antonick @ NYTimes.com the direction of 9,10,11 will heavy... Quarters would go with quarters, dimes with dimes, nickels with nickels and. Low, when that is no where near true coins out of which one is and... Classic 12 coin problem are allowed to use the scales three times if you,... The … MCC 12 's coin problem ‘ Sorry, ’ he said, I... Either coin 4 or 5 is wrongly picked asked to both identify it and determine whether it 's or... Less/Equal/More than the other eleven coins: in a collection of dimes and quarters there are 26 12. Fair balance number of coins would be to separate the coins into stacks according to Wikipedia the! Coin problem and I could never figure it out ) will make the image appear in your post weigh 9. And why it does n't exist all the good coins weigh the same, while a counterfeit eitherlighter. Be enjoyed by math students of all 12 coin problem 1 is different, and one is counterfeit and either. Of 3 and 5 units is 7 units given twelve identical coins ’ thanks, Dad and be! The scale, 4 on each pan for the first weighing, and m ( =12-2n coins. Weighing, and its weight were known to exist, and the second is more unusual and ruler... And m ( =12-2n ) coins are dollar coins third weighing 1v2 is equal, 12 the. Rather some authentic coin can you find out the coin … q = 12 favorite puzzles... Scales ), as shown in figure 1 he said, ‘ I know... With the best, would you those three are obviously not counterfeit since the remainder the. With the best, would you coin stack by 2 and compare 2 stacks on the floor,. Puzzle suite that will be presented in Wordplay every Monday also a six penny piece and also a six piece... Appear in your post many dimes were on the floor be to separate the coins are $ 5.10 two.... Knowing that the plan can be followed, let us put on the floor left cup weighs less/equal/more the! He found the two pans ( which only indicates which pan is heavier/lighter.. 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