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It reports that the Spring problems were more difficult than usual, with only a few perfect scores, and contrasts regular problems with a bonus challenge. The Summer solutions provide fully worked answers to multiple math puzzles, including score-table constraints, base-11 cryptic additions, and probability computations, as well as coin partition reasoning and complex-number limits.",{"@graph":69,"@context":122},[70,84,105],{"@type":71,"itemListElement":72},"BreadcrumbList",[73,77,79,82],{"item":74,"name":75,"@type":76,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":78,"name":9,"@type":76,"position":14},"https://docshare.wps.com/document/",{"item":80,"name":20,"@type":76,"position":81},"https://docshare.wps.com/document/exam/",3,{"item":83,"name":65,"@type":76,"position":19},"https://docshare.wps.com/document/brain-ticklers-spring-2011-solutions/269517/",{"url":83,"name":65,"@type":85,"image":86,"author":91,"headline":65,"publisher":94,"fileFormat":97,"inLanguage":63,"description":67,"dateModified":98,"datePublished":99,"encodingFormat":97,"isAccessibleForFree":100,"interactionStatistic":101},"DigitalDocument",{"url":87,"@type":88,"width":89,"height":90},"https://docshare.wps.com/thumbnails/brain-ticklers-spring-2011-solutions/269517.png","ImageObject",300,407,{"name":92,"@type":93},"Xiajie","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-10-20","2026-09-15",true,{"@type":102,"interactionType":103,"userInteractionCount":24},"InteractionCounter",{"@type":104},"ViewAction",{"@type":106,"mainEntity":107},"FAQPage",[108,114,118],{"name":109,"@type":110,"acceptedAnswer":111},"How were the Spring 2011 problems rated in difficulty and scoring?","Question",{"text":112,"@type":113},"The Spring Ticklers were more difficult than usual, and only a few scores were perfect. Two regular problems had lower correct-answer percentages than the Bonus problem.","Answer",{"name":115,"@type":110,"acceptedAnswer":116},"What is the decoded result of the cryptic additions in Problem 2?",{"text":117,"@type":113},"In base 11, the additions decode to USA + FDR = WPA and USA + WPA = PARK, with the digit 'a' representing 10.",{"name":119,"@type":110,"acceptedAnswer":120},"What method is used to solve Problem 4 about dividing 26 coins with equal heads?",{"text":121,"@type":113},"First split into a 10-coin group and a 16-coin group, then flip all ten coins in the 10-coin group. If H is the number of heads initially in the 10-coin group, the flipped result makes the number of heads match the 16-coin group.","https://schema.org",{"og:url":83,"og:type":124,"og:title":65,"og:site_name":95,"og:description":67},"article",{"robots":126,"canonical":83},"index,follow",{"doc_id":128,"site_id":62},269517,1791489059,{"code":4,"msg":5,"data":131},{"doc_id":128,"user_id":132,"nickname":92,"user_avatar":133,"doc_module":4,"category_id":19,"category_name":20,"doc_title":65,"doc_description":67,"doc_content":134,"file_id":135,"file_url":136,"file_type":137,"file_size":138,"view_count":24,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":81,"language":139,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":140,"faqs":141,"seo_title":142,"seo_description":67,"update_tm":143,"read_time":39},8814010472675,"https://avatar.qwps.com/avatar/WGlhamll","Brain Ticklers  \nRESULTS FROM  \nSPRING 2011  \nPerfect  \nCouillard, J. Gregory IL A’89  \n*Kimsey, David B. AL A’71  \n*Mayer, Michael A. IL A’89 Slegel, Timothy J. PA A’80  \n*Spong, Robert N. UT A’58  \n*Stout, Roger P. AZ B ’77  \n*Strong, Michael D. PA A’84  \nOther  \nAlexander, Jay A. IL G ’86  \nAron, Gert IA B ’58  \nBachmann, David E. MO B ’72  \n*Beaudet, Paul R. Father of member Bernacki, Stephen E. MA A’70 Bertrand, Richard M. WI B ’73 Brule, John D. MI B ’49 Conway, David B. TX I ’79  \n*Fenstermacher, T. Edward MD B ’80 Glaser, Anton Non-member Grewal, Rashi NJ G ’09 Harter, Eamonn T. ID G ’06  \n*Harvey, Arthur J. OH A’83  \n*Jones, John F. WI A’59 Oliver, Christopher R. AL E ’08 Prince, Lawrence R. CT B’91 Quintana, Juan S. OH Q’62 Rasbold, J. Charles OH A’83 Robillard, David J. MD G ’88  \n*Routh, Andre G. FL B ’89  \n*Schmidt, V. Hugo WAB’51 Sentman, Mark H. CA E ’86 Sentman, Andrew Son of member Silver, Robert E. NY R ’80 Stadlin, Walter O. NJ G ’52 Stribling, Jeffrey R. CA A’92 Stumpf, Kenneth A. MI D ’73 Summerfield, Steven L. MO G ’85 Sutor, David C. Son of member Van Wyk, Rogell IN A’59  \n*Voellinger, Edward J. Non-member Wills, Kenneth J. MT A’84  \n* Denotes correct bonus solution  \nSPRING REVIEW  \nThe Spring Ticklers appear to have been more difficult than usual, with only a few perfect scores. Two regular problems (Number 2 about a spider chasing an ant and Number 4 about card arrangements) had a lower percentage of correct answers than the Bonus problem.  \nSUMMER SOLUTIONS  \nReaders’ entries for the Summer problems will be acknowledged in the Winter Bent. Meanwhile, here are the answers:  \n1 The points awarded to Al, Bob, Carl, Don, and Ed on each question of the five-question math test are shown in the following table.  \nTo arrive at this table, note that the totals of the rows and the columns can only be expressed as a sum of the numbers 10, 7, 2 and 0; possibilities are: 9 = 7 + 2; 14 = 10 + 2 + 2 = 7 + 7; 19 = 10 + 7 + 2; 22 = 10 + 10 + 2; 29 = 10 + 10 + 7 + 2; 31 = 10 + 7 + 7 + 7 = 10 + 10 + 7 + 2 + 2; 34 = 10 + 10 + 10 + 2 + 2 = 10 + 10 + 7 + 7; 42 = 10 + 10 + 10 + 10 + 2. Because Q4 was so easy and Ed only got a total of 9 points, Ed must have been the one who got a 2 on Q4; therefore, Column 4 must be (10, 10, 10, 10, 2). Since Ed scored higher than Bob on Q5, Row 5 must be (0, 0, 0, 2, 7). Note that if the answer to Q2 were 17, then three 10s would be awarded, the sum of which is larger than 29; therefore, the second column must be (10, ?, 10, ?, 0) with the unknowns being 2 and 7.  \nNote that if the answer to Q3 were 5, then Bob would have scored two 10s, which sum to more than 19, so ‘5’is not the answer to Q3. Since Row 3 sums to 22, the correct answer must be ‘11,’ and Column 3 is (10, ?, ?, 10, 0) with unknowns 0 and 2. Since Row 1 sums to 34 and already contains three 10s, the other two scores must be 2, so Row 1 is (2, 10, 10, 10, 2). Column 1 sums to 14 and must contain a 10, but the 10 can’t be in Rows 2 or 3 because the row sums won’t work. Therefore, Don got the 10 on Q1.  \nNext, note that for Column 1 to sum to 14 and Row 3 to sum to 31, Carl must have scored a 2 on Q1. For Column 3 to sum to 22 and Row 3 to sum to 31, Carl must have scored a 2 on Q3. This means that Carl scored a 7 on Q5. Since Ed got more credit for Q5 than Bob, Bob must have scored a 2 on Q5 and a 7 on Q2. Finally, the only remaining score to be determined is Don’s score on Q2, which must be a 2.  \n2 Stated in base 11, the simultaneous cryptic additions decode as:  \nUSA + FDR = WPA : 5a4 + 320 = 914 USA + WPA = PARK : 5a4 + 914 = 1408,  \nwhere‘a’ stands for the digit 10.  \n3 In a game of 7-card stud with deuces wild, the probability of getting 5-of-a-kind is 7,622/1,672,307 ≈ 0.004558. There are 52!/(45!7!) = 133,784,560 possible 7-card hands. Consider the probability of getting one deuce and four cards of the same denomination. There are 12 possibilities for the 5-ofa-kind card (J, Q, K, A, or 3 through 10). There are four p","cbCaiqYCSiUC66Gz","https://ap.wps.com/l/cbCaiqYCSiUC66Gz","pdf",211760,"English","# Spring Review\n## Summer Solutions\n## Problem 1\n## Problem 2\n## Problem 3\n## Problem 4\n## Problem 5","[{\"question\":\"How were the Spring 2011 problems rated in difficulty and scoring?\",\"answer\":\"The Spring Ticklers were more difficult than usual, and only a few scores were perfect. Two regular problems had lower correct-answer percentages than the Bonus problem.\"},{\"question\":\"What is the decoded result of the cryptic additions in Problem 2?\",\"answer\":\"In base 11, the additions decode to USA + FDR = WPA and USA + WPA = PARK, with the digit 'a' representing 10.\"},{\"question\":\"What method is used to solve Problem 4 about dividing 26 coins with equal heads?\",\"answer\":\"First split into a 10-coin group and a 16-coin group, then flip all ten coins in the 10-coin group. If H is the number of heads initially in the 10-coin group, the flipped result makes the number of heads match the 16-coin group.\"}]","Brain Ticklers - Spring 2011 - Solutions | PDF",1789452910]