There were some great answers, but they all seem a little harder than my solution, which requires only two pieces of knowledge. 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> 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. at 15:01. Weigh coins 1-4 against 5-8. If 1,2,3,4 v 5,6,7,8 is not equal, mark which way each side moved. There is one other constraint: The balance is only capable handling an equal number of coins in the two pans. Jan’s cos(72) equals radical 5 minus 1 over 2 was great, but who knows that without looking something up? While written for adults, 722 722 778 778 778 778 778 584 778 722 722 722 722 667 667 611 278 333 556 556 556 556 280 556 333 737 370 556 584 333 737 552 The Problem: You are given twelve identical coins. Let’s assume that 1,2,5 went down again. This MCC had the second-highest unmultiplied coins, only behind MCC 7. Lessons Lessons. 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. 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. Without a reference coin. 556 556 556 556 556 556 889 556 556 556 556 556 278 278 278 278 << 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! His response: ‘I have no idea what you are talking about. The 12 marbles appear to be identical. Divide the coins into 3 groups: , , . 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. The bal… One can do comparison one by one and compare all 12 coins. How Do I Find and Operate Across Lite in Windows 8? There are 24 different potential answers: any of the 12 coins could be the fake, and the fake could be either heavier or lighter. Let’s assume 1,2,3,4 went down and 5,6,7,8 went up. Weigh coin 9 against coin 10; if they balance, then coin 11 is heavier. Solution for the "12 Coins" Problem. Our challenges this week were suggested by Numberplay regular Stephan Peers, an investment banker from Lafayette, Calif. I figured it out in my mid-20s. The balance provides one ofthree possible indications: the right pan is heavier, or the pans arein balance, or the left pan is heavier. How many nickels and how many dimes were on the floor? There are 12 coins. 667 778 722 667 611 722 667 944 667 667 611 278 278 278 469 556 After working on it for weeks, I gave up and asked him for the answer. 333 556 556 500 556 556 278 556 556 222 222 500 222 833 556 556 And send your favorite family puzzles to gary.antonick@NYTimes.com. SOLUTION TO THE 12 COIN PROBLEM. So why was it so high? Clearly we can discard the option of dividing into two … Your task is to identify the unusual marble and discard it. q = 12. Quarters would go with quarters, dimes with dimes, nickels with nickels, and so on. How can you find odd coin, if any, in minimum trials, also determine whether defective coin is lighter or heavier, in the worst case? What happens if the balance is level? many of the concepts here are suitable for and can be enjoyed by math students of all ages. 1015 667 667 722 722 667 611 778 722 278 500 667 556 833 722 778 Can you determine the counterfeit in 3 weightings, and tell if it is heavier or lighter? Some of the coins may be left aside. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 750 750 The Twelve-Coin Problem. 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). (Sometimes the puzzle features billiard balls instead of coins, but the problem is the same.) And thank you, Mr. Peers. << �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�?ܿ^. Here is a fancy chart I made to illustrate my point. If a counterfeit coin were known to exist, and its weight were known, the amount of That’s really the same as mine. However, one is counterfeit and may may either lighter or heavier than the other eleven coins. 611 611 611 611 611 611 611 549 611 611 611 611 611 556 611 556 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. 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? We have no other information. Here is a rough statement of the puzzle: 1. That’s it for this week. There are two possibilities. Classic problem with 12 coins ( or marbles) one of which is fake. I assume you know how to drop perpendiculars and bisect lines. Nope! He posed the problem and I could never figure it out. endobj One is counterfeit and is either heavier or lighter than the other 11. You have 12 coins that appear identical. 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. Also — If you could train with the best, would you? 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. One of them is slightly heavier or lighter than the others. Can you find the counterfeit coin in three weighings? Determine the conterfeit coin and whether it is light or heavy in three weighings using a balance scale. Each time we use the balance, we have three different possible answers: the left cup weighs less/equal/more than the right cup. 1�φ8��n�?6)pє�� 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. You have 12 coins that all look exactly the same. If the value of the coins is $1.95, how may of each type do they have? Her books, “It's Not Jack and Betty have 28 coins that are nickels and dimes. So 40 years after, I told my Dad that I solved the problem. The third weighing is 9v10. I can check to make sure this works: 14×$1 + 12×$0.25 = $14 + $3 = $17. 556 556 556 556 556 556 556 556 556 556 333 333 584 584 584 611 Weigh coin 12 against coin 1 to determine whether coin 12 is heavier or lighter. Show Solution. The first is a classic we ran in similar form years ago, and the second is more unusual and a If there is $29.65 overall, how many of each are there? 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. We should not underestimate the challenge, both from a technical point of view and in terms of design. 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. 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). Somewhere else was a passing statement on the fact that the ratio of the long arms on a star to the base was the golden ratio. If scale remains balanced after first weigh: Second Weigh A as follows. It indicates that the faulty coin must lie among the m coins left aside. So the problem changes to m coins and two measurements. The N is 12 cents. Either they balance, or they don't. How can you tell even with 2 coins at the end which is the odd one out? ] You aregiven twelve coins. 400 549 333 333 333 576 537 278 333 333 365 556 834 834 834 611 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. lot of fun. 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. 556 556 556 556 556 556 556 549 611 556 556 556 556 500 556 500 Rounds of silver destined to be coins, at the Old Glory Mint in Utah. 14 0 obj There are 12 more nickels than dimes. For instance, if both coins 1 and 2 are counterfeit, either coin 4 or 5 is wrongly picked. One of them is fake. Can you determine the counterfeit in 3 weightings, and tell if it is heavier or lighter? One way to bring some order to the mess of coins would be to separate the coins into stacks according to their value. one of them is counterfeit. q = 12. /Creator (easyPDF SDK 6.0) … Construct a perfect pentagon with a compass and a ruler. I hope it wasn’t the only result of my graduate education.”, “Same Dad issue: great problem, no solution. 8 0 obj You are allowed to weigh only three times. Perspectives from Olympians Gwen Jorgensen and Clark Hewitt’s Enhancer (see note at the bottom of this post) will make the image appear in your post. 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. 2. How can you find out which one is fake using only 3 weighings on a balance? Any change of coins on either side of the scale is considered to be a weighing. Note that the unusual marble may be heavier or lighter than the others. How do you want to group them? Our second challenge involves a bit of geometry. Along with discussion about the day's challenge, you'll Gary 12 coin problem. If they balance, then the different marble is in the group 9,10,11,12. 667 778 722 667 611 722 667 944 667 667 611 333 278 333 584 556 Fake coin assumed to be lighter than real one. You have a classical balance with two pans (which only indicates which pan is heavier/lighter). 2. There are 12 coins. See below for the construction. 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-�� The puzzles, “My Dad proposed the coin puzzle when I was, like, 10 years old. 750 278 278 500 500 350 556 1000 333 1000 556 333 944 750 500 667 >> With a balance beam scale, isolate the counterfeit coin in three moves. [ >> 2 0 obj 278 333 556 556 556 556 260 556 333 737 370 556 584 333 737 552 ��t��(�?���zy��X����!�T�4x[��D@��)�S\L������ ���:& #���-�Fk�h��� �ې3�@��IQZt�2��2~�VHq�e��a-]:���! Having scales to compare coins (or marbles). In fact, 11 of them are identical, and one is of a different weight. 667 667 667 667 667 667 1000 722 667 667 667 667 278 278 278 278 That means that either 1 or 2 is counterfeit and it is heavy. Solvers Solvers. 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). 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. Solution. the two plate ones, with no precise v Let’s begin with the perfect pentagon. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 750 750 556 556 333 500 278 556 500 722 500 500 500 334 260 334 584 750 Here's an old silver three penny piece and also a six penny piece. /ModDate (D:20120426205302-07'00') [ Example: In a collection of dimes and quarters there are 6 more dimes than quarters. With help from the mnemonic "ma do like me to find fake coin," three weighings will automatically determine which coin is fake. The total value of the coins are $5.10. over 100 logic and math puzzles for The New York Times, secretly believes every math problem can be solved using circles and straight lines. Whether it is the heavier or lighter one? If coins 0 and 13 are deleted from these weighings they give one generic solution to the 12-coin problem. Welcome to our conversation about word games. Those three are obviously not counterfeit since the counterfeit will always cause the scale to move the same way. I’m sure I have a misunderstanding here. Classic problem with 12 coins ( or marbles) one of which is fake. A harder and more general problem is: At some point I learned that the ratio of the the long side to the short side of a star is in the Golden Ratio. All appear to be identical, as far as you cantell by eye, but you are told that one of them is a counterfeit. %���� It may weigh more or less than a real one. One of them is fake: it is either lighter or heavier than a normal coin. 556 750 222 556 333 1000 556 556 333 1000 667 333 1000 750 611 750 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. The second weighing is 1,2,5 v 3,4,6 where three of the coins change sides. I see a lot of people saying that this MCC's coins were super low, when that is no where near true. First step: 1-9 weight with 10-18, A. balance ==> fake ball in 19-27. Strange Symbols Find the fake coin and tell if it is lighter or heavier by using a balance the minimum number of times possible. << In other words, 12 of the coins are quarters. Discussion. Numberplay is a puzzle suite that will be presented in Wordplay every Monday. Then either one of the following situations will occur: (A): or (B): or (C): If (A) occurs, then one of the must be fake. B. 722 722 778 778 778 778 778 584 778 722 722 722 722 667 667 611 /Filter /FlateDecode 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, if on the third weighing 1v2 is equal, then 6 is the counterfeit and it is light. 611 611 389 556 333 611 556 778 556 556 500 389 280 389 584 750 If there is $29.65 overall, how many of each are there? �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���;? Coin Word Problems Examples: 1. endobj 400 549 333 333 333 576 556 278 333 333 365 556 834 834 834 611 Not til I read Mario Livio’s book The Golden Ratio. It has been presented in many different ways. 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. I had to explore a lot more to be able to figure out a strategy. /Pages 4 0 R We’ll start by leaning into —. Which is divide the coin stack by 2 and compare 2 stacks on the scales. 12 coin problem. Word Problems: Coins Word. If they do not balance, then the coin that weighs more is the heavier coin. For example, the largest amount that cannot be obtained using only coins of 3 and 5 units is 7 units. 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. With help from the mnemonic "ma do like me to find fake coin," three weighings will automatically determine which coin is fake. 333 556 611 556 611 556 333 611 611 278 278 556 278 889 611 611 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 12 Coin Balance Problem Answer. The New York Times’ weekly puzzle blog Numberplay has moved to a new and improved location. ���h��g����d�&�`k"sX��#[sX�����!����\����q.T��.�_���~S��o:WiZܷȁZ�Z�k#4!�G�S�J���(�ypz�ӱ(�hhũ E\�� � 3 0 obj get backstage insights about puzzlemaking and occasional notes from The Times's puzzlemaster, Will Shortz. Thank you, Andy, and giant thanks to Stephan Peers for this week’s challenges. endobj For the 12 Identical Balls Problem, using the same method, the maximum number of balls can be up to 27. The nickels and dimes all fell on the floor. For those who don’t know about dynamic programming it is according to Wikipedia, The counterfeit weigh less or more than the other coins. Fake coin assumed to be lighter than real one. Having scales to compare coins (or marbles). Gary Antonick, who has created or edited 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? Fabulously interesting. 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. Hi, You have 12 coins. ‘Sorry,’ he said, ‘I just know the problem, not the solution.’ Thanks, Dad. I first read this problem in a book of short stories by Ethan Canin called “The Palace Thief.” This was in the second story. Bi-set or tri-set? I think I was reading Mario Livio’s book The Golden Ratio. It hasn't been solved I am afraid I asked Mr. Peers where he found the two problems. Discussion Solution For solutions, mail me or post a comment. Allgenuine coins have the same mass, while a counterfeit is eitherlighter or heavier than a genuine coin. 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…. This is now the complete answer to the 12 coin problem. '”, Let’s give these a try. 12 Coin Problem There are 12 identical coins. I put the two together and was thrilled. 11 are identical and 1 is different (different weight). More efficiently one can do it using Decrease By Factor algorithm. Was it because of Parkour Tag? The harder task is educating the coin … Burckle. If equal, 11 is counterfeit. 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. You have 3 weighings of a scale (i.e. Problem 4: (The classic 12 coin puzzle) You are given two pan fair balance. 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. 278 278 355 556 556 889 667 191 333 333 389 584 278 333 278 278 you have 12 coins. If two coins are counterfeit, this procedure, in general, does not pick either of these, but rather some authentic coin. the counterfeit coin is either slightly heavier or slightly lighter than all the others. In other words, 12 of the coins are quarters. 12 coins problem This problem is originally stated as: You have a balance scale and 12 coins, 1 of which is counterfeit. 12 coins problem This problem is originally stated as: You have a balance scale and 12 coins, 1 of which is counterfeit. Thank you as well to everyone who participated in our discussion: Technic Ally, John W., Donald Quixote, Ramona D’Souza, D-Ferg, So that the plan can be followed, let us number the marbles from 1 to 12. Hence, by using the balance twice, we can anticipate 9 different outcomes; and by using it three times — 27 different outcomes. You are only allowed 3 weighings on a two-pan balance and must also determine if the counterfeit coin is heavy or light. Step 1: Weigh against . Weigh 1 v 2 and which ever one goes down again is the counterfeit coin. The counterfeit weigh less or more than the other coins. I guess he got a kick out of things like that. 556 556 556 556 556 556 556 556 556 556 278 278 584 584 584 556 Thanks for all the comments. Since the answer works in the original exercise, it must be right. Here’s Mr. Peers: The Pentagon Problem gave me trouble for years. 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. problem solving. If instead the set 9,10,11 is *heavier* than 1,2,3, then any one of coins 9,10,11 could be heavier. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 975 722 722 722 722 667 611 778 722 278 556 722 611 833 722 778 sB�p�6�"���[����D%3�f(g:����bE���gg���6,s�wVGa9�5t da�S��~� Weigh coins 1-4,9 and 5-8,10 . C. 1-9 heavier ==> fake ball in 10-18. But it’s quite possible to make it easy for users to keep custody of their keys, combining high security with great UX. This one's a (great) classic. MCC 12's coin problem and why it doesn't exist. One is counterfeit and is either heavier or lighter than the other 11. Let us say we have n coins on each pan for the first weighing, and m (=12-2n) coins are lying aside. Since the remainder of the twenty-six coins are dollar coins, then there are 26 – 12 = 14 dollar coins. One of the coins is counterfeit. First Weigh. 12 Coins. One can do comparison one by one and compare all 12 coins. Since the answer works in the original exercise, it must be right. }~ORȘC��M�Q%�~zo�۲�!����d�6�0SF�bJd�ݾ�������U���j��f�p=*�o������;#�73L�\����-���?�+'A��N You are asked to both identify it and determine whether it is heavy or light. /CreationDate (D:20120426205302-07'00') endobj However, one is counterfeit and may may either lighter or heavier than the other eleven coins. Second step: take the group of nine balls (A is 19-27, B is 1-9, C is 10-18) weight the first six balls with three on each side. edited 5 years ago. A harder and more general problem is: For some given n > 1, there are (3^n - 3)/2 coins, 1 of which is counterfeit. You have 12 identically looking coins out of which one coin may be lighter or heavier. 556 556 556 556 556 556 889 500 556 556 556 556 278 278 278 278 You have 12 coins that appear identical. Great stuff that quadratic equation solution. You are allowed to use the scales three times if you wish, but no more. I have to add something to my answer. I began with 8 coins on the scale, 4 on each side. You have 12 coins and a 2-pan balance scale. 11 0 obj /Producer (BCL easyPDF 6.02 \(0342\)) 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. Can you find out the coin which is different, and also whether it's heavier or lighter? /Type /Catalog If not equal, the direction of 9,10,11 will determine heavy or light. Thanks, Ravi, for knowing that the diagonal of a pentagon is the Golden Ratio. If equal, 12 is the counterfeit and weigh it against any other coin to determine if it’s heavy or light. For the first weighing let us put on the left pan marbles 1,2,3,4 and on the right pan marbles 5,6,7,8. Example: In a collection of dimes and quarters there are 6 more dimes than quarters. /Length 2892 stream Let's number the coins 1-12. 1-9 lighter ==> fake ball in 1-9. Since the remainder of the twenty-six coins are dollar coins, then there are 26 – 12 = 14 dollar coins. %PDF-1.3 722 722 722 722 722 722 1000 722 667 667 667 667 278 278 278 278 Since we are only required to handle 12 coins, things are still on track. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 You have 12 coins that all look exactly the same. endobj 556 750 278 556 500 1000 556 556 333 1000 667 333 1000 750 611 750 Show Step-by-step Solutions More efficiently one can do it using Decrease By Factor algorithm. To share a picture of your solution, just upload to an image-hosting site (like imgr) and include the link in your comment. Readers weighed in with a variety of solutions, including this by Andy, which identified the counterfeit Bernard's AllExperts page.. 12 Coin problem. Determine the conterfeit coin and whether it is light or heavy in three weighings using a balance scale. Wow, so fancy! 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? Our second challenge this week was the classic Twelve Coin Problem. ] 1 0 obj Show Step-by-step Solutions 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. Here are the detailed conditions: 1) All 12 coins look identical. 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). x��Z[�ۺ~�_�G-�UŻ�hn'�K[t�>$}�n�[�uj{�ȿ��PIIkSN�����X��p�qn���?��qv���rY����Z��0�u� �����q��4� �붹Y_^ �����d�F6�0�B�F96�w��f� ��`$�z�_�.��!r����/��5K�����jǖ��g��9��g�@�l�=֫�w�v͖�k�����ņ-h�����ԹӶ`W���|h�-���&�~ �5l��|a ���g�w�[�t�\H�g��a��|���n>�2�N�Ն�����l�n��"Ev[���D���ʩ�v�m����!V$q�c�;a�}�6��սd���SfB��0cx.�UQ�E���]���� ]�J,��휧�j��ބ��R�*3�V*^�'���O`DO�e^������N�UU�9�0���6Ue������@�ǀm����K���-�.Z���)��U8�ߡ�T�8j*�E��Y���R��+�W-���oX�do�0��2�7[6�u�H��n�X�W �q��s{{��}�\YJ�}�����%I�*W�FA�7[� ��T����u?��K)���o���)e�W��ʴ�m��/��p8p^�JvU f�������a9U�ۡ�1dPN7��8,��!O�K��Ii����+J���d���I�uI�D��B��9OZJ?��(c08&���-GM�9B���\2B�;�x�6�(:��*��O��I���xx�d�N��p?75�Fw��p{���V�C(D�;�N�r�;��)Z�i���>��0�����ה�f�lks1��I+A����}�������3 l�z_o7��V�.-yAY��dcI� Here you'll find a new blog post for each day's crossword plus a bonus post for the Variety puzzle. 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. all the good coins weigh the same, while the counterfeit one weights either more or less than a good coin. Therefore, we can't do the … You are provided an equal-arm balance (sometimes called a scaleor scales), as shown in figure 1. >> P.M.S., It's You” and “Create Your Life Lists” are available where all fine literature is sold. He is a visiting scholar at Stanford University, where he studies mathematical 278 333 474 556 556 889 722 238 333 333 389 584 278 333 278 278 I can check to make sure this works: 14×$1 + 12×$0.25 = $14 + $3 = $17. 8���μ�D���>%�ʂӱA氌�=&Oi������1f�Ė���g�}aq����{?���\��^ġD�VId݆�j�s�V�j��R��6$�����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 , nickels with nickels, and the second weighing is 1,2,5 v 3,4,6 where three of the coins $... 6 more dimes than quarters balance the minimum number of coins 9,10,11 could be heavier or lighter the... 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