Punjab Technical University, Jalandhar B. Pharmacy
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Le téléchargement nécessite un accès à la bibliothèque YouScribe Tout savoir sur nos offres
- mémoire - matière potentielle : storage
- mémoire
- mémoire - matière potentielle : cells
- dissertation
- exposé
- exposé - matière potentielle : packages
- indefinite integrals of standard forms
- calcium as calcium oxalate
- magnesium as magnesium pyrophosphate
- inorganic analysis
- function of a function
- plant tissue cultures as sources of drugs
- aluminium as aluminium oxide
- pharmaceutical analysis
- pharmaceutical chemistry
- plant cell
Challenges of the Week Solutions Spring Semester 2005-2006 Challenge of the Week # 1 - January 13 to January 20: Insert arithmetical signs ( chosen from+,,,) and parentheses in the following to make a true equation. 1 11 = 2006 2 6 6018
Solved by Holly Bertram, Mark Bingham, Jillian Centers, Doug Cichon, Natalie Cox, Christina Dockus, Kari Donoho, Katie Dzielski, Katherine Fehrenbacher, Je Flaximan, Kyle Frey, Katie Harris, Sarah Kistler, Lisa Krause, Kelli Larsen, Jackie Malan, Bryan Martin, Matt Niemerg, Michael Ortel, Nicole Petrey, Shannon Price, Stephen Puricelli, Drew Roberts, Jill Seelhoefer, Symon Shmukler, Shannon Smith, Nicole Srutowski, Lucas Sweitzer, Nicole Szaykovics, Angie Taylor, Kirsten Thomas, Elise Vendegna, Amy Vitzthum, Anthony Westendorf, Melissa Younker, and David Zoerb.Partially solved by Amanda Truttman.Other submissions from Dan Davis, James Eiter, Amy Helpingstine, Charles Hamlin, Nicole Marczewski, Andy McGilliard, Ryan Moran, and Lisa Pendola. „ « 1 11 = 2006 2 66018 Challenge of the Week # 2 - January 20 to January 27:A box contains jewels. There are less than 1000 jewels in the box.If 2/9 of the jewels are diamonds, 4/11 are rubies, 1/7 are sapphires, and the rest are emeralds; how many emeralds are there in the box?Justify your answer. Complete solutions received from Evan Bright, Kati Dzielski, Katie Harris, Lisa Krause, Jackie Malan, Andy McGilliard, Matt Niemerg, Michael Ortel, Shannon Price, Kael Rewers, Shannon Smith, Jill Sparenberg, Angie Taylor, and Kristen Thomas.Partial solutions from Kelley Altmeyer, Brian Anderson, Holly Bertram, Mark Bingham, Doug Cichon, Natalie Cox, John DeGiulio, Kari Donoho, Je Flaxman,Ivan Hargrove, Andrew Hockley, Steve Keepes-England, Sarah Kistler, Kelli Larsen, Nicole Marczewski, Bryan Martin, Lisa Pendola, Drew Roberts, Nicole Scutowski, Jill Seelhoefer, Amanda Truttmann, and Amy Vitzhum.Other submissions from Katie Fehrenbacher, Jillian Centers, Jon Holmgren, Kelley Mullaney, and Nicole Szajhovics.Since the number of jewels and the number of jewels of each type are integers, the number of jewels must be divisible by 7, 9 and 11.Since these are relatively prime numbers, the number of jewels must be divisible by 693.But the number of jewels is less than a thousand and there is only one positive multiple of 693 that is less than 1000.Hence the number of jewels is 693, the number of diamonds is 154, the number of rubies is 252, and the number of sapphires is 99.This means that the number of emeralds is 188. Challenge of the Week # 3 - January 27 to February 3: A certain safe can be opened by entering the correct 7 digit code.Each of the digits in the code is either a two or a three, the number of twos in the code is more than the number of threes in the code, and the code is a number which is divisible by 6.Find all 7 digit codes that will open the safe. Justify your answer. The best solutions were received from Holly Bertram, Evan Bright , Jillian Centers, Doug Cichon, Katie Harris, Steve Keepes-England, Kelli Larsen, Jackie Malan, J.J. Murphy, Michael Ortel, Shannon Price , Julie Reeve, Jill Sparenberg, Nicole Srotowski, Kristen Thomas, and Elise Vendegna.Partially correct solutions from Mark Bingham, Natalie Cox, Brandon Dawson, Kari Donoho, Katie Dzielski, Amy Helpingstine, Jen Holmgren, Sarah Kistler, Lisa Krause, Patrick Lohan, Bryan Martin, Andy McGilliard, Matt Niemerg, Nicole Petrey, Michelle Range, Jill Seelhoefer, Symon Shmuckler, Shannon Smith, Nicole Szajhovics, Angie Taylor, Amanda Truttmann, Tara Welderman.Other submission from Katie Fehrenbacher.Since the number is divisible by three, the sum of the digits must be divisible by 3. Therefore, there must be zero, three or six digits of two.Since the number of twos is more than the number of threes, there must be six twos in the code.The last digit must be even, since 2 divides the code.There are six possible codes 3222222, 2322222, 2232222, 2223222, 2222322, and 2222232.
Challenges of the Week - Solutions - Spring Semester 2005-2006
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Challenge of the Week # 4 - February 3 to February 10: You have a plate and four napkins.The plate and the napkinsareallintheshapeofanequilateraltriangle,andtheyallhavethesamesize.Eachnapkinhasaxed,but unknown, orientation.It is not allowed to rotate a napkin, but a napkin may be translated in any direction.Show that it is always possible to translate the napkins so that the plate is completely covered by napkins.
No correct solutions were received.The most nearly complete solutions which were submitted by Kari Donoho and Elise Vendegna.Other submissions from Mark Bingham, Katherine Fehrenbacher, Evan Hargrove, Katie Harris, Steve Keepes-England, Lisa Krause, Kelley Maline, Nicole Petrey, Jill Seehoefer, Jessica St.Pierre, and Amanda Truttman.This week’s award will added to the next Challenge.The following diagram shows an equilateral triangle which has been divided into four congruent equilateral triangles and then a circle has been circumscribed about each of the small triangles.All circles are congruent since all triangles are congruent.The circumscribed circle about the “middle” small triangle is tangent to the sides of the large triangle and is thus the same size as the inscribed circle of the large triangle.
Now, the plate and the napkins are the same size.The plate is covered by four of the small circles.Each of the small circles determined by the plate can be covered by a translate of a napkin by moving the napkin so that the inscribed circle of the napkin coincides with the small circle from the plate. Each of the small circles covers one of the small triangles on the corner of the plate.In addition, each small circle covers one-third of the central small triangle.Thus, three small circles cover the plate.By the argument above, the plate can be covered by the translates of three napkins. ChallengeoftheWeek#5-February13toFebruary24:Inasetofnineconsecutiveintegers,therstisdivisible by 10, the second is divisible by 9, the third is divisible by 8, the fourth is divisible by 7, and so on until the last which is divisible by 2. How many such sets of consecutive integers are there?What are the numbers in each set? The best solution to this week’s Challenge was given by Elise Vendegna.Other solutions were provided by Mark Bingham, Katie Fehrenbacker, Katie Harris, Amy Helpingstine, Lisa Krause, Andy McGilliard, Michael Ortel, Shannon Price, Jill Seelhoefer, Shannon Smith, Nicole Szaykovics, Angie Taylor, Kristen Thomas, and Amanda Truttman.Other submissions from Holly Bertram, Brett He erkamp,EvanHargrove,KelleyMullaney,andJessicaPierre.First consider the case that the sequence of integers is increasing.LetN10 be the rstnumber. WewantN10 to be divisible by 10,N9 to be divisible by 9,N. . ,8 to be divisible by 8, .NTherefore2 to be divisible by 2.Nmust be divisible by 10,9,8, . . . ,2. Since
Challenges of the Week - Solutions - Spring Semester 2005-2006
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the least common multiple of 10,9,8, . . .2 is 2520 we must have thatN= 2520kfor some integerk. Thenumbers in this case are: 2520k10,2520k9,2520k8, . . . ,2520k2. Ifthesequenceofnumbersisdecreasing,lettingtherstnumberbeN+ 10an argument similar to that given above showsN= 2520kfor some integerkand the numbers are: 2520k+ 10,2520k+ 9,2520k+ 8, . . . ,2520k+ 2. Challenge of the Week # 6 - February 24 to March 3: Ifais a real number, theintegral part ofa, denoted[a], is the largest integer less than or equal toa. Thefractional part ofa, denoted{a}, is that portion of the number that is to right of the decimal point whenaThusis written as a decimal.[3.14] = 3and{2.71828}= 0.71828. Find all positive real numbers,x, y, z, which satisfy the following system of equations: x+ [y] +{z}= 12.91 y+ [z] +{x}= 11.43 z+ [x] +{y}= 13.76
Solved by Evan Bright, Katie Dzielski, Amy Helpingstine, Andy McGilliard, Michael Ortel, Shan-non Price, Symon Schmulker, Nicole Szaykovics, and Angie Taylor.Other submissions from Evan Hargrove, Katie Harris, Sarah Kistler, Jill Seelhoefer and Jessica St.Pierre.Note that for a positive real numberwwe havew={w}+ [wthe three equations together and using this observation gives that]. Adding 2x+ 2y+ 2z= 12.91 + 11.43 + 13.76 = 38.10. Hencex+y+z= 19.ldsubtr05.Stiacthngrsequteoitaorfnihtmeiys y[y] +z {z}= 6.14 ={y}+ [z]. Since [z] is an integer and{y}is a decimal, greater than or equal to 0, but less than one, it follows that [z] = 6 and {y}= 0.14. Similar using the second equation and then the third equation give{z}= 0.62, [x] = 7, and{x}= 0.29, [y] = 5, respectively. Hence,x= 7.29,y= 5.14, andz= 6.62. Challenge of the Week # 7 - March 3 to March 10: Five congruent circles are situated as indicated in the following diagram. The pointPis the center of the leftmost circle.
P
Show how to draw a single straight line which passes throughPand such that half of the area inside the circles is on one side of the line and half of the area inside the circles is on the other side. Complete solutions were submitted by Holly Bertram, Kari Donoho, Katie Harris, Shannon Price, and Jill Sparenberg.Partial solutions were submitted by Mark Bingham, Justin Capps, Katie Fehren-bacher, Evan Hargrove, Amy Helpingstine, Kaitlin Lewis, Kelley Maline, Nicole Marczewski, Andy McGilliard, Amy Moran, Ryan Moran, Michael Ortel, Lisa Pendola, Nicole Petrey, Jill Seelhoe-fer, Shannon Smith, Jessica St.Pierre, Nicole Szajkovics, Angie Taylor, Kristen Thomas, Amanda Truttman, Elise Vendegna, Amy Vitzthum, Anthony Westendorf and Courtney Wilson.Other sub-missions from Ty Bartholomew and Sarah KistlerConsider the line passing throughPand the point where the common tangents to the circle, illustrated below:
Challenges of the Week - Solutions - Spring Semester 2005-2006
P
4
The circle whose center isPis divided in half by this line, because the line goes through the center of that circle. The other four circles are divided in half because the line goes through a point of central symmetry of those four circles. Challenge of the Week # 8 - March 24 to March 31: Put the numbers 1,2,3,4,5,6,7,8,9, one per circle and without repetition in the circles of the following diagram so that the sum of the numbers on each side of the triangle equals 20 and the product of the numbers on the corners is odd.
Give all nonequivalent solutions where two solutions are said to be equivalent if one can be obtained from the other by a symmetry of the triangle and/or by interchanging the two numbers along one side of the triangle. Completely solved by Mark Bingham, Michael Ortel, Shannon Price, Jill Sparenberg, Jessica St. Pierre, Amy Vitzhum, Anthony Westendorf.Partially solved by Kelly Altmeyer, Adam Ahlers, Dan Davis, Katie Dzielski, Amy Helpingstine, Steve Keepes-England, Sarah Kistler, Lisa Krause, Kelly Maline, Jill Seelhoefer, Shannon Smith, Nicole Srutowski, Nicole Szaykovics, Angie Taylor, Kirsten Thomas, Amanda Truttman, and Courtney Wilson.Other submissions from Rachel Bailey and Ryan Meyers.The sum of the numbers along each edge is 20.Thus, if the numbers along all three sides are added together the sum will be 60.This will equal the sum of all of the numbers (45) plus the sum of the numbers on the three corners (which is 15).Because the product of the numbers on the corners is odd, all numbers at the corners must be odd.It is easy to see that the three numbers on the corners are either 3, 5, and 7 or 1, 5, and 9.If the corner numbers are 1, 5 and 9, then 8 cannot be on the same side as 9.Hence along one side of the triangle we have 1, 8, 5 and 20 - 1 - 8 - 5 = 6. The two numbers along the side containing 1 and 9 must sum to 10.Since no number is used twice in the triangle, the two numbers along this side must be 3 and 7, since the other cases — 1 and 9, 2 and 8, 4 and 6 — are impossible.By elimination, the remaining side of the triangle contains 5, 2, 4, and 9.Suppose the corner numbers are 3, 5 and 7.Consider the side containing 3 and 5.The sum of the other two numbers must be 12 and they must be 9 and 3, 8 and 4, or 7 and 5.Since numbers are used only once, it must be that the numbers 3, 5, 8 and 4 are along one side of the triangle.The side containing 5 and 7 cannot contain 9, for then the sum would exceed 20.Hence, 9 is on the side with 3 and 9.The fourth number on this side is 1.Finally, the numbers on the thirdsideofthetrianglewillbe7,2,6,and5.Thetwonon-equivalentcongurationsareshownbelow 5 7
8
6 41 2
2
9
6
3 7 98 4 5 1 3
Challenges of the Week - Solutions - Spring Semester 2005-2006
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Challenge of the Week # 9 - March 31 to April 7: This is the annual April Fool’s version of the Challenge of the Week 1. Whydoes a Swiss barber prefer to cut the hair of two French men rather than the hair of one German man? 2. Thereare some shoes on the table.Some are black, some are brown, some are for the left foot, some are for therightfoot.Isitalwayspossiblethattherearetwoshoesofdi erentcolor,oneofwhichisfortheleftfoot and one of which is for the right foot? 3. Showhow to divide a thick, circular cake into 8 parts with three straight-line cuts?The diagram below shows the top view of how a cake can be cut into seven parts with three straight line cuts.
m n 4. Supposem < nor ?. Which fraction is closer to 1, n m For each question, justify your answer. Partial solutions were received from Mark Bingham, Dan Davis, Kari Donoho, Katie Fehrenbacher, Evan Hargrove, Kristen Kennard, Sarah Kistler, Kaitlin Kasper, Lisa Krause, Lisa McDaniel, Andy McGilliard, Michael Ortel, Jill Seelhoefer, Shannon Price, Amanda Truttman, Amy Vitzthum, and Anthony Westendorf 1. Becauseyou get more money from two haircuts than from one haircut. 2.Yes!Bya“specialpair”wemeantwoshoes—oneleft,oneright—ofdi erentcolors.Pickupanyleftshoe. Ifanyoftherightshoeshaveadi erentcolor,thenthereisaspecialpair.Ifalloftherightshoeshavethe samecolorasthechosenleftshoe,thentheshoeofadi erentcolor,whichexistsbyassumption,mustbea left shoe.This shoe and any right shoe are a special pair. 3. Makeone cut horizontal to the bottom, halfway down.This gives two congruent cakes.Before removing the top “cake”, divide both cakes into four equal parts with two cuts. m mn n mn mn mn nm 4. Now,|1|=| |, while|1|=| |. Sincem < n,| |>| |. Hence|1|>|1|. n nm mm nm n n m Therefore isfarther away from 1 than. Butthis is April Fool’s.The above argument only works ifm m n n andnConsider what happens ifare positive.m=2 andnIn fact, if= 1.|m|<|n|, thenis farther away m m from 1 than, as the above argument shows. n Challenge of the Week # 10 - April 7 to April 14: You are at the shore of lake and need to measure out 1 gallon of water.All you have is a red 5 gallon bucket and a blue bucket which contains either 3 or 4 gallons, but you can’t remember which. What procedure can you use to measure out a single gallon. (Of course, there are no markings on either bucket.) The best solutions were provided by Holly Bertram, Megan Goeckner, Andrew Mounce, Michael Ortel, and Shannon Price.Other correct solutions were provided by Steve Keepes-England, Andy McGilliard, Jill Seelhoefer, Jill Sparenberg, and Angie Taylor.Partial solved by Katie Fehrenbacher, Kaitlin Kasper, and Elise Vendegna.Other submissions from Kari Donoho, Sarah Kistler, Lisa Krause, Lisa McDaniel, Amy Moran, Ryan Moran, Shannon Smith, and Nicole Szajokovics.Take the red bucket andllitup.Fillupthebluebucketfromtheredbucketandemptythebluebucketbackintothelake.Then empty the red bucket into the blue bucket.There is either 2 or 1 gallons of water in the blue bucket, depending upon whether there the blue bucket holds three or four gallons. Nowlluptheredbucketand“top-up”thebluebucketfromtheredbucket.Thenemptythebluebucketback into the lake.There will be either 4 or 2 gallons in the red bucket, depending upon whether the blue bucket holds threeorfourgallons.Finally,llupthebluebucketfromtheredbucket.Ifthereisanywaterleftintheredbucket, the blue bucket must hold three gallons and there is exactly one gallon left in the red bucket.If there is no water left in the red bucket, the blue bucket must hold four gallons and there are two gallons in the blue bucket.In this lastcase,theeasiestwaytomeasureoutonegallonistoemptythebluebucket,lluptheredbucketfromthelake andthenpourfourgallonsintothebluebucket(llingitup).Onegallonofwaterwillremainintheredbucket.
Challenges of the Week - Solutions - Spring Semester 2005-2006
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Challenge of the Week # 11 - April 14 to April 21: In the representation ofA, shown below, there aremoccurences of 9 andnoccurences of 8, wheremandnare positive integers. 09 81 9 8 9 8 9 8 @ A A= 98
Asmandnrange over all positive integers, what are values of the remainders whenAis divided by 7? Justify your answers. 3 3 27 Note: Be careful about the notation here. For example,3means3(= 7,625,597,484,987).Don’t confuse this 3 33 with(3 )which is27( = 19,683). Lisa Krause, Matt Niemerg, Nicole Srutowski submitted papersWhenn= 1, the remainder is 1, when n >1, the remainder is 0. For any value ofm, since 8 is one more than a multiple of 7, the number 8 8 8 8 8 is one more than a multiple of 7.Clearly, whenn= 1, since 9 is two more than a multiple of 7,Awill be one more than a multiple of 7. 9 9 9 9 33 Ifn >. But1, then 3 divides the exponent 99 =(7 + 2)is eight more than a multiple of 7 and hence 1 more than a multiple of 7.Therefore, the remainder is 0 in this case.
