2014 State Competition Solutions
2014 State Competition SolutionsAre you wondering how we could have possibly thought that a Mathlete would be able toanswer a particular Sprint Round problem without a calculator?Are you wondering how we could have possibly thought that a Mathlete would be able toanswer a particular Target Round problem in less 3 minutes?Are you wondering how we could have possibly thought that a Mathlete would be able toanswer a particular Team Round problem with less that 10 sheets of scratch paper?The following pages provide solutions to the Sprint, Target and Team Rounds of the2014 MATHCOUNTS State Competition. Though these solutions provide creative andconcise ways of solving the problems from the competition, there are certainly numerousother solutions that also lead to the correct answer, and may even be more creative or moreconcise! We encourage you to find numerous solutions and representations for theseMATHCOUNTS problems.Special thanks to volunteer author Mady Bauer for sharing these solutions with us and therest of the MATHCOUNTS community!
2014 State CompetitionSprint Round1. A mouse weighs 25 grams and a dogweighs 5000 grams. The weight of the5000 200 times the weight of thedog is25mouse. 200 Ans.2. π₯π₯ 3 4First the positive value:π₯π₯ 3 4π₯π₯ 7Now the negative value: (π₯π₯ 3) 4 π₯π₯ 3 4π₯π₯ 1The two solutions are 7 and -1.The product of the solutions is:7 1 7 Ans.3. A square target has a side length of 16.The smaller square has a side length of 4.Find the probability that the dart lands inthe smaller square.The area of the smaller square is:4 4 16The area of the larger square is:16 16 256The probability that the dart lands in thesmaller square is:161 Ans.256both the same, i.e. 60 .Each section denoted by those angleshas an area of 1/6 the area of the circlefor a total of 1/3 the area of the circle.Given that the diameter of the circle is 2,then the radius is 1.The area of the circle is Οr2 Ο. 1/3 thearea of the circle is (1/3) Ο. Therefore,k 1/3 Ans.ππ(π₯π₯ 1)5. ππ(π₯π₯) 2ππ(2) 4Find the value of ππ(10) ππ(7)From the first equation:ππ(π₯π₯ 1) 2ππ(π₯π₯)ππ(3) 2ππ(2) 8ππ(4) 2ππ(3) 16ππ(5) 2ππ(4) 32Anyone see what the value of f(x) really is?ππ(π₯π₯) 2π₯π₯Therefore, ππ(7) 27 128 and ππ(10) 210 1024ππ(10) ππ(7) 1024 128 896 Ans.6. If n is the number of seconds in a day, findthe largest prime factor of n.The number of seconds in the day is:24 60 60 (2 2 2 3) (2 2 3 5) (2 2 3 5)Clearly, 5 is the largest prime.5 Ans.7. How many convex polygons with an evennumber of sides are in the figure shown?164. The diameter of Circle P is 2. Themeasure of MPN is 120 . The area ofthe shaded region is kΟ. Find k.Given that the measure of MPN is 120 ,then that is also the measure of QPR.120 120 240360 β 240 120The measure of MPQ and NPR areA convex polygon has all angles less than180 . A polygon must have at least 3sides but we are only dealing with 4 sidesso letβs start with polygons of size 4. Thefirst 4-sided shape we can look at is arhombus. There are 6 of those.
Next we have the trapezoid. There are 6of those:Now 6 sided polygons. We have someβirregularβ shaped polygons, like this one.But remember that we are looking forconvex polygons. Itβs obvious that wehave an angle that is larger than 180 .This doesnβt qualify.So what is left, is the original hexagon.6 6 1 13 Ans.8. Stage 1 has the number 1. Obviously thesum is 1.Stage 2 has the numbers 2, 3 and 1, 2.The sum is 8.Stage 3 has the numbers, 3, 4, 5 and 2, 3,4 and 1, 2, 3. The sum is 12 9 6 27.Stage 4 has the numbers 4, 5, 6, 7 and 3,4, 5, 6 and 2, 3, 4, 5 and 1, 2, 3, 4 for asum of 22 18 14 10 64.See the pattern? The sum of the numbersin a stage? If the stage is n, then the sumof the numbers is ππ3 .The sum of the numbers for stage 7 is73 343. Ans.9. The sequence 1, 3, 4, 7, 11, 18, 29 isconstructed by adding the two previousterms (after the first two terms). Howmany of the first 100 terms of thissequence are multiples of 5?So far, none. Letβs go a little further butwe only need to put down the oneβscolumn.7, 6, 3, 9, 2, 1, 3, 4, 7, 1, 8 β wait aminute. Weβre into repeating the onescolumn. We never got the value 0 or 5.Therefore, there will be 0 terms in thissequence that are multiples of 5.0 Ans.10. 3 different gifts were bought for 3children. The gifts were wrapped aftertheir names were put on them. Find theprobability that no child opens a giftlabeled with his or her name.Name the gifts 1, 2, and 3.Name the kids A, B and C.A was meant to have 1, B was meant tohave 2 and C was meant to have 3.If A gets present 2, then B can get 3 andC can get 1.If A gets present 3, then B can get 1 andC can get 2. Thatβs 2 choices. And thosesame choices apply if we start with B orC. I.e.,A 2, B 3, C 1 andA 3, B 1, C 2.There are 3! 6 combinations.2/6 1/3 Ans.11. The sides of 3 similar regular pentagonsare in the ratio of 2:5:7. The area of thesmallest pentagon is 40. Find the area ofthe largest pentagon.The formula for the area of a regular5pentagon is 2 π π π π where s is the side of thepentagon and a is the apothem (a linedrawn from the center of the pentagon tothe midpoint of a side).Let 2x the side of the smallestpentagon. Then the apothem for thepentagon is:
5540 π π π π 2π₯π₯π₯π₯ 5π₯π₯π₯π₯228 π₯π₯π₯π₯8ππ π₯π₯Since the ratio of the smallest to themiddle pentagon is 2:5, the side of themiddle pentagon is 5x and the apothem ofthe middle pentagon is:8 5 40 20 π₯π₯π₯π₯ 2 2π₯π₯Finally, the ratio of the middle to thelargest pentagon is 5:7.Therefore, the side of the largest is 7x.The apothem of the largest pentagon is:20 7 28 π₯π₯π₯π₯ 5The area of the largest pentagon is:28 55 7π₯π₯ 7 28 5 7 14 π₯π₯22490 Ans.12. Find the area of the quadrilateral withvertices (0,0), (1,2), (3,4) and (6,5).Letβs look at a graph of the quadrilateral.Itβs going to be tough to try and find thearea of the quadrilateral directly, so letβsconsider the pentagon with vertices (0,0),(1,2), (3,4), (6,5) and (6, 0). As shown,the area of the quadrilateral is the area ofthe pentagon minus the area of the righttriangle with vertices (0, 0), (6, 5) and(6, 0).To determine the area of the pentagon,letβs partition it into two trapezoids and aright triangle, as shown.Therefore, the area of the quadrilateral is[(1/2) 1 2] [((2 4)/2) 2] [((4 5)/2) 2] β [(1/2) 6 5] 1 6 27/2 β 15 (14 27 β 30)/2 11/2 5 1/2 Ans.13. The integers a, b, c, d and e form anarithmetic sequence. Their sum is 440.Find the largest possible value of e.A sequence is an arithmetic sequence ifthe terms differ by the same value.Let x that difference. Thenππ ππ π₯π₯ππ ππ 2π₯π₯ππ ππ 3π₯π₯ππ ππ 4π₯π₯ππ ππ ππ ππ ππ ππ ππ π₯π₯ ππ 2π₯π₯ ππ 3π₯π₯ ππ 4π₯π₯ 5ππ 10π₯π₯ 440ππ 2π₯π₯ 88To find the largest value of e, a must be assmall as possible. The numbers in theseries are non-negative. Suppose ππ 0.0 2π₯π₯ 882π₯π₯ 88π₯π₯ 44ππ ππ 4π₯π₯ 0 (4 44) 176 Ans.14. β789XYZβ is a 6-digit integer consistingof six distinct digits. It is divisible by 7, 8and 9. Find the three-digit integer βXYZβ.A number that is divisible by 9 must havethe sum of the digits add up to a valuethat is divisible by 9.7 8 9 24X Y Z must be divisible by 3 but notdivisible by 9, otherwise the entire numberwill not be divisble by 9.We have combinations of{1, 2, 3} (sums to 6){1, 2, 9}, {1, 3, 8}, {1, 4, 7}, {1, 5, 6},
{2, 3, 7}, {2, 4, 6}, {3, 4, 5} (sums to 12){3, 4, 8}, {3, 5, 7}, {4, 5, 6} (sums to 15){5, 7, 9}, {6, 7, 8} (sums to 21){7, 8, 9} (sums to 24)But we canβt use any groups that have a7, 8 or 9 in them because all numbers aredistinct. That takes us down to:{1, 2, 3}{1, 5, 6}{2, 4, 6}{3, 4, 5}{4, 5, 6}A number is divisible by 8 if its last threedigits are divisble by 8. Obviously, thelast digit must also be an even number.So we can look at each of thesecombinations choose an even number tobe the ones digit and see if the three digitnumber is divisible by 8.{1, 2, 3} gives us 132 and 312. 312 isdivisible by 8.{1, 5, 6} gives us 156 and 516. Neitherare divisible by 8.{2, 4, 6} gives us 246, 264, 426, 462,624 and 642. 264 and 624 are divisibleby 8.{3, 4, 5} gives us 354 and 534. Neitherare divisible by 8.{4, 5, 6} gives us 456, 546, 564 and 654.564 is divisble by 8.So we are left at looking at each of789,312789,264789,624789,564Only 789,264 is divisble by 7 as well.264 Ans.15. The roots of the equationπ₯π₯ 2 6π₯π₯ ππ 0 are in the ratio of 2:1.Find k.If we looked at π₯π₯ 2 6π₯π₯ 9 0, the tworoots are x 3. No ratio of 2:1 there.But if we looked at(π₯π₯ 2)(π₯π₯ 4) π₯π₯ 2 6π₯π₯ 8 0the roots are -2 and -4 which do satisfythe requirement that the roots are in theratio of 2:1. Therefore, k 8. Ans.16. A cone is sliced by planes parallel to itsbase into 3 pieces of equal height. Findwhat fraction of the original volume is inthe middle piece.Let h the height of the entire cone.Let x the length of the height betweenthe top of the cone and the first plane,between the first and second planes andbetween the second and third planes.Then h 3xIf r the radius of the smallest cone, then2r and 3r are the radii of the other twocones since the sides of the cone and theplanes all form similar triangles.1The volume of a cone is 3 Οππ 2 β.The volume of the entire cone is1Ο (3ππ)2 3π₯π₯ 9Οππ 2 π₯π₯3The volume of the smallest cone is11Ο (ππ)2 π₯π₯ Οππ 2 π₯π₯33The volume of the middle cone is81Ο (2ππ)2 2β Οππ 2 π₯π₯33The volume of the middle piece is thevolume of the middle cone minus thevolume of the smallest cone.78 2 1 2Οππ Οππ π₯π₯ Οππ 2 π₯π₯333The fraction of the original volume is7 27Οππ π₯π₯73 3 29Οππ π₯π₯9 277Ans.2717. A six-sided die has two red, two blue andtwo yellow faces. The die is rolled 3times. Find the probability of getting eachcolor once.The probability of getting a particular colorwhen the die is rolled is2 1 6 3
The probability of getting a certain colorthe first time (say red), a different color(say blue) the second time and a differentcolor again (say yellow) the third time is:11 1 1 3 3 3 27There are 3! 6 different combinations ofred, blue and yellow. Therefore, theprobability of getting each color once is:62 Ans.27918. A code has to include 4 distinct digits inthe range of 1 through 5. The positivedifference between any two consecutivedigits must be at least 2. Find how manycodes can be used.Start with 1.135214251 5 2 4 β Thatβs 3 possibilities.Now start with 2.24132415251325142 5 3 1 β Thatβs 5 possibilities.Now start with 3:3142315235143 5 2 4 β Thatβs 4 possibilities.Now start with 4:41354152415342514 2 5 3 β Thatβs 5 possibilities.And finally, start with 5514252415 3 1 4 β Thatβs 3 possibilities.Thatβs a total of 3 5 4 5 3 20 combinations.20 Ans.19. The composition, to the nearest wholepercent of a class is 43% male and 57%female. What is the minimum number ofstudents that could be in the class?Nearest whole percent is the operativeterm here.Clearly, if we have 2 students, one maleand one female, itβs 50% for either.123 students would be or . That wonβt33work.4 students obviously wonβt work either norwould 5 students or 6 students.What about 7 students?3 0, .4285 43%77 Ans.20. Express 2710 3145 in base 5.I donβt mind adding in a base differentfrom 10 but multiplication is nasty. Letβsconvert 3145 into base 10, perform themultiplication and then turn it back into avalue in base 5.3145 (3 52 ) (1 51 ) (4 50 ) 75 5 4 8484 27 22682268 1875 375 15 3 (625 3) (125 3) (5 3) (1 3) (3 54 ) (3 53 ) (3 51 ) (3 50 ) 330335 Ans.21. π₯π₯ 2 1π₯π₯ 2 3, π₯π₯ 01Find the value of π₯π₯ π₯π₯1 21 π₯π₯ π₯π₯ 2 2 2π₯π₯π₯π₯1 2 π₯π₯ 3 2 5π₯π₯1π₯π₯ 5 Ans.π₯π₯22. A bag contains red marbles and bluemarbles. The probability that two marbleschosen are red is 1/5. The probability thattwo marbles chosen are blue is 1/5.Determine the number of marbles in thebag.Let x the number of red marbles.x is also the number of blue marbles.π₯π₯ 11π₯π₯ 2π₯π₯ 2π₯π₯ 1 5π₯π₯ 2 π₯π₯1 24π₯π₯ 2π₯π₯ 5
5π₯π₯ 2 5π₯π₯ 4π₯π₯ 2 2π₯π₯π₯π₯ 2 3π₯π₯ 0π₯π₯(π₯π₯ 3) 0x cannot be 0 since we have to haveβsomeβ marbles here.π₯π₯ 32x is the number of marbles or6 Ans.23. Triangles ABD and DEF are isoscelesright triangles. A, D, F and C are collinearand B, E and C are collinear.AB BD 4. ED EF 2. Find thelength of AC.Since DBE is congruent to FEC,triangles ABC and DEC are similar (AA).Therefore, AB/DE AC/DC 4/2 (4 2 2 2AD FC)/(2 2 FC) 8 2 4 FC 8 2 4 2 2 FC 2 FC 4 2 FC 2 2 units. Thus,AC 4 2 2 2 2 2 8 2. Ans.24. How many digits are in the integerrepresentation of 230 ?Looking at the powers of 2, we have2 4 8 16 32 64 128 256 512 1024 20484096 8192 16384 32768The number of digits in each can bewritten as:1 1 1 2 2 2 3 3 3 etc.Therefore, the number of digits needed torepresent 230 is 10. Ans.25. The diagonals of a parallelogram are 82and 30. One altitude is 18. Find thesmallest possible area of theparallelogram.The area of a parallelogram is ππβ where ππis the base of the parallelogram and β isthe altitude. To find the area, we need tofind the base, b. Break that into twopieces, x and y, which are formed bydropping the altitude. The intersection ofthe diagonals also bisects the altitude.ππ π₯π₯ π¦π¦π¦π¦ 2 92 412π¦π¦ 81 1681π¦π¦ 2 1681 81 1600π¦π¦ 40π₯π₯ 2 92 152π₯π₯ 2 81 225π₯π₯ 2 225 81 144π₯π₯ 12ππ π₯π₯ π¦π¦ 12 40 52The area of the parallelogram is:ππβ 52 18 936But does this give us the smallestpossible area? Probably not since this isthe obvious solution.In the diagram above, the altitude wascompletely contains within theparallelogram. Is there a way in which thealtitude is not contained within theparallelogram?This happens when the two of the parallellines of the parallelogram areentirely divorced from each either β i.e,move DC down and move AB up and nopoint of AB will touch a point of DC.h EF and EG GF 9DB 30 and DG GB 15AC 82 and AG GC 41Let x AF. Thenπ₯π₯ 2 92 412π₯π₯ 2 81 1681π₯π₯ 2 1681 1600π₯π₯ 40Let y BF. Thenπ¦π¦ 2 92 152π¦π¦ 2 81 225π¦π¦ 2 225 81 144π¦π¦ 12AF AB BFπ₯π₯ π΄π΄π΄π΄ π¦π¦
Now, instead of adding x and y, we get tosubtract y from x.40 π΄π΄π΄π΄ 12π΄π΄π΄π΄ 40 12 28AB is the base so the area is28 18 504504 is definitely smaller than 936.504 Ans.26. 7 prizes are to be distributed between 3people with a guarantee that each persongets at least one prize. Find the number ofways the prizes can be distributed.Prizes can be given to the three people inthe following combinations:1, 1, 51, 2, 41, 3, 32, 2, 37!7 6For 1, 1, 5 there are 5!2! 2 21 waysfor the 5 gifts to be given to one particularperson. To make this moreunderstandable, assume person #3 hasthe 5 gifts. Then person #1 gets gift Aand person #2 gets gift B (A and B arenot part of the 5) or person #1 gets gift Band person #2 gets gift A. So, for one setof 5 gifts going to person #3, there aretwo combinations for persons #1 and #2.Therefore, for 21 combinations of the 5gifts there are 21 2 42. This is justfor Person #3 receiving the 5 gifts. Thereare 42 more for Person #1 and Person#2 receiving the 5 gifts, respectively.42 3 1267!7 6 5For 1, 2, 4 there are 4!3! 3 2 1 35ways to choose the 4. Again, assume thatPerson #3 receives the 4 gifts. Assumethat Person #1 receives the single giftand Person #2 receives the two gifts.3!There are 2!1! 3 ways to allocate the 3gifts between Person #1 and Person #2.If we now let Person #1 receive two giftsand Person #2 receive the single gift,thatβs another 3 combinations for a total of6 combinations while Person #3 receivethe 4 gifts. 35 6 210 combinationswhen Person #3 receives the 4 gifts.Adding in combinations when Person #1receives the 4 gifts and Person #2receives the 4 gifts we get a total of210 3 630 combinations.For 1, 3, 3, letβs have Person #3 receivethe single gift. There are6!6 5 4 20 ways to allocate the3!3!3!first group of 3 prizes, say, to Person #1.But allocation of 3 prizes to Person #1also allocates the other 3 prizes to Person#2. This allocation covers allcombinations for both Person #1 andPerson #2 (e.g., Person #3 gets present1, Person #1 gets presents 2-4 andPerson #2 gets 5-7 or Person #1 getspresents 5-7 and Person #2 gets 2-4etc.) Therefore, we have 20 combinationsper single present for Person #3 for atotal of 7 20 combinations when Person#3 gets the single present. This ends upbeing a total of 140 3 420combinations when Person #1 andperson #2 get the single present.Finally, for 2, 2, 3, there are 35 ways to4!4 3allocate the group of 3 and 2!2! 2 6ways to allocate the sets of 2 presents tothe other two people.35 6 210 ways and 210 3 630total combinations.The final total of combinations is 126 630 420 630 1806 ways Ans.27. The fourth degree polynomial equation ofπ₯π₯ 4 7π₯π₯ 3 4π₯π₯ 2 7π₯π₯ 4 0 has 4roots, a, b, c and d. Find the value of thesum1 1 1 1 ππ ππ ππ ππ4π₯π₯ 7π₯π₯ 3 4π₯π₯ 2 7π₯π₯ 4 is of the form:ππ4 π₯π₯ 4 ππ3 π₯π₯ 3 ππ2 π₯π₯ 2 ππ1 π₯π₯1 ππ0 whereππ4 1, ππ3 7, ππ2 4, ππ1 7 andππ0 4The quickest way to solve this is tounderstand what the product of(π₯π₯ ππ) (π₯π₯ ππ) (π₯π₯ ππ) (π₯π₯ ππ)looks like.
(π₯π₯ ππ) (π₯π₯ ππ) (π₯π₯ ππ) (π₯π₯ ππ) π₯π₯ 4 π₯π₯ 3 (ππ ππ ππ ππ) π₯π₯ 2 (ππππ ππππ ππππ ππππ ππππ ππππ) π₯π₯(ππππππ ππππππ ππππππ ππππππ) ππππππππππ4 1ππ3 7 ππ ππ ππ ππNote that this is the sum of the roots.ππ2 4 ππππ ππππ ππππ ππππ ππππ ππππNote that this is the sum of the product oftwo of the roots.ππ1 7 ππππππ ππππππ ππππππ ππππππNote that this is the sum of the product ofthree of the roots.Finally, ππ0 4 ππππππππThis is the product of all 4 roots.Now letβs look at1 1 1 1 ππ ππ ππ ππ1 1 1 1 ππ ππ ππ ππ ππππππππππ ππ ππ ππππππππ ππππππ ππππππ ππππππ οΏ½οΏ½π ππππππ ππππππ ππππππππ7 1 Ans.ππππππππππ0428. Find the remainder when 1112 is dividedby 13. One way to solve this problemwithout a calculator is by using modulararithmetic.Since 11 mod 13 β2 mod 13, we knowthat 1112 mod 13 (β2)12 mod 13. But(β2)12 212, so we just need to find theremainder when 212 is divided by 13.212 4096, and 13 315 4095. So,the remainder is 1 Ans.29. ππ(ππ) ππ2 1 if n is oddππππ(ππ) if n is even2For how many integers from 1 to 100does ππ ππ ππ(ππ) 1 for somenumber of applications of f?Letβs start with 1.ππ(1) 12 1 1 1 22ππ ππ(1) ππ(2) 12At this point, weβre going to repeat 2, 1, 2,1, etc.Now start with 2.2 12ππ(ππ(2) ππ(1) 2And again, weβll just repeat 1, 2.Now start with 3.ππ(3) 32 1 9 1 1010ππ(ππ3)) ππ(10) 52ππ ππ ππ(3) ππ(5) 52 1 2626 13ππ ππ ππ ππ(3) ππ(26) 2Continued application of the function f willonly result in larger and larger values.Therefore, 3 is not a part of this (and noneof the rest of the odd values).Now start with 4.4ππ(4) 222ππ ππ(4) ππ(2) 12So 4 works. We already know that 5 willnot so letβs look at 6.6ππ(6) 32ππ ππ(6) ππ(3) 10We now have enough information. Forn 1, 2, 4, we can have the situationwhere multiple applications of f will resultin the value 1. Odd values starting with 3will never result in the value of the functionbeing 1 because the value just getsbigger and bigger. That leaves multiplesof 4.8ππ(8) 42We already know that f(4) works so 8 isalso good.12ππ(12) 62And now we can see that any multiple of3 (and, of course, 4) cannot be used.Basically, what we are down to arepowers of 2. 1, 2, 4, 8, 16, 32, 647 Ans.ππ(2)
30. In pentagon ABCDE, E and C areright angles and m D 120, AB 12,AE BC 18 and FD DC. Find ED.Letβs start by constructing segment EC.Then weβll draw a perpendicular segmentfrom A that intersects EC at a point weβllcall F. Lastly, weβll draw anotherperpendicular from D that intersect
2014 MATHCOUNTS State Competition. Though these solutions provide creative and concise ways of solving the problems from the competition, there are certainly numerous other solutions that also lead to the correct answer, and may even be more creative or more concise! We encourage you to find numerous solutions and representations for these
In all, this means that competition policy in the financial sector is quite complex and can be hard to analyze. Empirical research on competition in the financial sector is also still at an early stage. The evidence nevertheless shows that factors driving competition and competition have been important aspects of recent financial sector .
He is the first prize winner of the National Flute Association's Young Artist Competition, WAMSO Minnesota Orchestra Competition, MTNA Young Artist Competition, Claude Monteux Flute Competition, second prize winner of the William C. Byrd Competition and finalist at the Concert Artists Guild Victor Elmaleh International Competition.
September: 2013 33,391.18 9/24/2013 October: 2013 33,391.18 10/24/2013 December: 2013 65,031.50 12/20/2013 January: 2014 33,099.37 1/23/2014 February: 2014 33,099.37 2/24/2014 March: 2014 33,099.37 3/24/2014 April: 2014 31,662.23 4/25/2014 May: 2014 31,662.23 5/22/2014 June: 2014 31,662.24 6/26/2014 392,881.03
Competition Statistics Friday, May 9, 2014 Orlando, FL 2014 Raytheon MATHCOUNTS National Competition National Competition 4.00 24.16 43.00 9.37 Indiv. Total Minimum Average
Preparation from Question Banks and Practice to Students 01.07.09 to 22.10.09 School level Quiz Competition 23.10.09 to 24.10.09 Cluster level Quiz Competition 17.11.09 to 20.11.09 Zonal level Quiz Competition 01.12.09 to 04.12.09 District level Quiz Competition 04.01.10 to 06.01.10 Region
competitions including the Columbus State Guitar Symposium, Guitare Lachine Competition (Montreal), Rosario Competition (Columbia, SC), Schadt String Competition (Allentown, PA), the Dr. Luis Sigall Competition (Vina del Mar, Chile), Guitar Foundation of America Competition, and the
The MATHCOUNTS Competition Series promotes mathematics achievement through a series of engaging "bee" style contests. Like the Spelling Bee, MATHCOUNTS is a program where students progress from the School Competition, to the Local/Chapter Competition, to the State Competition, and ο¬nally to the National Competition.
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