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for Elementary & Middle Schools Mathematical Olympiads November 18, 2014 Copyright © 2014 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved. 1A Time: 3 minutes !What!is!the!value!of!N!that!makes!the!sentence!true?!!1!+!2!+!3!+!4!+!5!+!6!=!3!×!N 1B Time: 5 minutes !Pascal's!Triangle!is!shown!here.!The!first!and!last!number!in!each!row!is!1.!Each!of!the!other!numbers!is!the!sum!of!the!two!numbers!diagonally!above!it,!as!shown!by!the!arrows.!Five!rows!are!shown.!If!the!pattern!is!continued!for!two!more!rows,!what!is!the!sum!of!all!seven!numbers!in!that!row?! 1C Time: 5 minutes !In!a!class!of!27!students,!16!like!video!games!and!20!like!cartoons.!If!12!students!like!both!video!games!and!cartoons,!how!many!students!do!not!like!either?!! 1D Time: 5 minutes !Two!9!cm!x!13!cm!rectangles!overlap!as!shown!to!!form!a!9!cm!x!22!cm!rectangle.!!What!is!the!area!!of!the!overlapping!rectangular!region? 1E Time: 7 minutes !The!following!three!statements!are!true:!!!!Δ!+!!Δ!!+!!◊!!=!18!!!!◊!!+!!◊!!+!⌂!=!22!!!⌂!+!⌂!+!!Δ!!=!17!What!is!the!value!of!!Δ!+!◊!+!⌂!?! Please fold over on line. Write answers on back. 1 11 121 1331 14641

for Elementary & Middle Schools Mathematical Olympiads November 18, 2014 Student Name and Answer Student Name and Answer Student Name and Answer Student Name and Answer Student Name and Answer N = Please fold over on line. Write answers in these boxes. 1A 1B 1C 1D 1E sq cm

Division Contest for Elementary & Middle Schools Mathematical Olympiads December 16, 2014 E 2 Copyright © 201 4 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved. 2A Time: 3 minutes What is the value of 9 + 26 + 83 + 55 + 45 + 17 + 74 + 91? 2B Time: 4 minutes The number 6 has exactly four unique factors: 1, 2, 3, and 6. How many counting numbers less than 20 have an odd number of unique factors? 2C Time: 5 minutes For the first half of the season, Alpha team won of their meets. For the second half of the season, they won of their meets. If both halves of the season had the same number of games, what is the fewest possible number of wins that they have? 2D Time: 7 minutes Sixteen 1cm by 1cm by 1cm cubes are glued together, face - to -face, as shown. The object is then entirely painted red. What is the total area, in sq cm, of all of the red painted surfaces? 2 E Time: 7 minutes The numbers from 1 through 9 are placed in the grid, exactly one per box without repeats. The numbers shown at the end of each row are the products of the numbers in that row. The numbers shown at the bottom of each column are the products of the numbers in th at column. What is the sum of the numbers in the four corners of the 3 by 3 grid? Please fold over on line. Write answers on back. 15 64 378

Division Contest for Elementary & Middle Schools Mathematical Olympiads December 16, 2014 E 2 Student Name and Answer Student Name and Answer Student Name and Answer Student Name and Answer Student Name and Answer Please fold over on line. Write answers in these boxes. 2A 2B 2C 2D 2E sq cm

for Elementary & Middle Schools Mathematical Olympiads January 13, 2015 Copyright © 2014 by MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved. 3A Time: 4 minutes !Compute:!100!+!91!–!82!+!73!–!64!+!55!–!46!+!37!–!28!+!19!–!10.! 3B Time: 5 minutes !What!is! ! 1 3 !of! ! 2 4 !of! ! 3 5 !of! ! 4 6 of!3000?!!3C Time: 5 minutes !In!the!figure!shown,!the!“H”!has!been!formed!by!removing!two!!2!x!4!rectangles!from!the!top!middle!and!bottom!middle!of!a!!6!x!10!rectangle.!!The!“H”!is!to!be!completely!tiled!with!1!x!1!tiles,!!which!come!in!boxes!of!6.!What!is!the!fewest!number!of!boxes!of!!tiles!that!must!be!bought!to!tile!the!“H”?!3D Time: 7 minutes !The!girls!on!a!softball!team!are!sharing!a!bag!of!fresh!strawberries.!!If!every!girl!has!5!whole!strawberries,!there!are!3!left!over.!!If,!instead,!the!girls!decide!to!share!the!strawberries!evenly!among!themselves!and!their!four!coaches,!and!everyone!takes!4!whole!strawberries!each,!there!are!none!left!over.!How!many!girls!are!on!the!softball!team?!3E Time: 7 minutes !The!tenPdigit!number!3872649A0B!is!divisible!by!36.!The!letters!A!and!B!each!represent!single!digit!even!numbers.!Find!the!sum!A!+!B.! Please fold over on line. Write answers on back.

for Elementary & Middle Schools Mathematical Olympiads January 13, 2015 Student Name and Answer Student Name and Answer Student Name and Answer Student Name and Answer Student Name and Answer Please fold over on line. Write answers in these boxes. 3A 3B 3C 3D 3E

for Elementary & Middle Schools Mathematical Olympiads February 10, 2015 Copyright © 2014 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved. 4A Time: 3 minutes !Pablo!has!nickels!and!dimes!and!no!other!coins.!He!has!five!more!dimes!than!nickels.!!Altogether,!he!has!$1.40.!!How!many!nickels!does!Pablo!have? 4B Time: 4 minutes !What!is!the!value!of!(18!×!40)!+!(12!×!40)!+!(20!×!28)!+!(20!×!12)? 4C Time: 5 minutes !What!number!between!200!and!300!is!exactly!divisible!by!3,!by!5,!and!by!7? 4D Time: 7 minutes !The!sum!of!the!digits!of!the!number!789!is!24.!!How!many!3Ndigit!numbers!have!the!sum!of!their!digits!equal!to!24!including!789? 4E Time: 7 minutes !A!pyramid!of!1cm!×!1cm!×!1cm!cubes!is!formed.!The!bottom!layer!has!a!9!×!9!arrangement!of!these!cubes.!The!second!layer!has!a!7!×!7!arrangement.!The!third!layer!has!a!5!×!5!arrangement.!The!fourth!layer!has!a!3!×!3!arrangement.!The!top!layer!is!a!single!cube.!The!sides!and!tops!of!the!pyramid!were!painted!(not!the!bottom).!How!many!square!centimeters!in!total!were!painted?! Please fold over on line. Write answers on back.

for Elementary & Middle Schools Mathematical Olympiads February 10, 2015 Student Name and Answer Student Name and Answer Student Name and Answer Student Name and Answer Student Name and Answer Please fold over on line. Write answers in these boxes. 4A 4B 4C 4D 4E sq cm

for Elementary & Middle Schools Mathematical Olympiads March 3, 2015 Copyright © 2014 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved. 5A Time: 3 minutes !Add:!102!+!203!+!304!+!405!+!506!+!607!+!708!+!809!+!901. 5B Time: 4 minutes !In!each!box,!the!three!numbers!on!the!top!row!are!used!to!obtain!the!number!in!the!bottom!row.!!If!the!same!pattern!is!used!in!each!box,!!what!is!the!value!of!N?! 5C Time: 6 minutes !Tracy!has!A#quarters!and!B!dimes!with!a!total!value!of!$3.45.!Tracy!has!more!quarters!than!dimes.!How!many!different!values!of!A!can!Tracy!have?! 5D Time: 7 minutes !Jimmy!is!filling!up!a!pool!using!a!large!hose!and!a!small!hose.!!The!large!hose,!working!alone,!could!fill!the!pool!in!3!hours.!!The!small!hose,!working!alone,!could!fill!the!pool!in!5!hours.!!The!small!hose!is!turned!on!and!allowed!to!run!for!an!hour.!!Then!the!large!hose!is!turned!on,!and!both!run!until!the!pool!is!full.!!How!many!hours!did!it!take!to!fill!the!pool?! 5E Time: 7 minutes !Square!ABCD!is!composed!of!36!squares!of!the!same!size,!as!shown.!The!area!of!square!ABCD!is!180!square!centimeters.!What!is!the!area,!in!square!centimeters,!of!the!shaded!region?! Please fold over on line. Write answers on back. 4 11 3 2 5 26 1 5 2 N 3 7 2 13 7 3

for Elementary & Middle Schools Mathematical Olympiads March 3, 2015 Student Name and Answer Student Name and Answer Student Name and Answer Student Name and Answer Student Name and Answer Please fold over on line. Write answers in these boxes. 5A 5B 5C 5D 5E sq cm

for Elementary & Middle Schools Mathematical Olympiads November 18, 2014 Copyright © 2014 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved. 1A METHOD 1: Strategy: Use grouping. Observe!that!the!six!numbers!being!added!are!consecutive!and!can!be!grouped!into!three!pairs!that!will!have!the!same!sum.!Starting!with!the!two!middle!numbers:!!3!+!4!=!7;!and!working!outwards:!2!+!5!=!7!and!1!+!6!=!7.!This!makes!three!groups!of!7.!The!sum!is!equal!to!3!×!N,!therefore!N!=!7.! METHOD 2: Strategy: Use the commutative property and grouping. Observe!that!3!×!N!means!3!groups!of!N!or!equivalently!N!groups!of!3.!Count!the!number!of!groups!of!3.!One!group!of!3!is!1!+!2,!3!alone!is!a!second!group!of!3,!4!+!5!is!three!more!groups!of!3!and!6!is!two!more!groups!of!3.!A!total!of!7!groups!of!3!so!N!=!7.!!!!! METHOD 3: Strategy: Use arithmetic and algebra. Add!the!numbers!on!the!left!side!of!the!equation!and!then!divide!both!sides!of!the!equation!by!3.!Thus!21!=!3!×!N,!so!N!=!7.! FOLLOW-UP: The average of a set of consecutive odd integers is 36. If the sum of all the integers in this set is 288, what is the greatest integer in the set? [43] 1B METHOD 1: Strategy: Find a pattern for the sums of the numbers in each row. The!sums!of!the!numbers!in!each!row!starting!at!the!top!of!the!triangle!are:!1,!2,!4,!8,!and!16.!Each!subsequent!row!is!double!the!previous!row!so!the!sixth!row!adds!to!32!and!the!seventh!row!has!a!sum!of!64.$ METHOD 2: Strategy: Continue the diagram and then add. The!sixth!row!is!1!!5!!10!!10!!5!!1!and!the!seventh!row!is!1!!6!!15!!20!!15!!6!!1.!The!sum!of!the!numbers!in!the!seventh!row!is!64. METHOD 3: Strategy: Use the diagram to determine the pattern. Note!that!from!the!top!to!the!bottom!of!the!diagram!every!row!has!two!arrows!from!each!number,!representing!the!fact!that!in!the!row!below,!each!of!the!numbers!in!the!row!above!will!be!counted!twice.!Thus!the!sum!in!each!succeeding!row!is!twice!the!sum!of!the!previous!row.!The!sixth!and!seventh!rows!add!to!32!and!64,!respectively.! FOLLOW-UP: The sum of the numbers in the thirteenth row is 4,096. If the sum of the numbers in row P is subtracted from the sum of the numbers in the thirteenth row, the difference is 3,840. What row does P represent? [9] N = 7 64 3 36 sq cm 19 SOLUTIONS AND ANSWERS 1A 1B 1C 1D 1E = + = + = + =

1C METHOD 1: Strategy: Create a Venn diagram. !Start!the!diagram!by!placing!the!12!in!the!region!that!displays!students!!who!like!both!video!games!and!cartoons.!Since!16!students!like!video!!games,!16!–!12!=!4!students!like!only!video!games.!Since!20!students!like!cartoons,!20!–!12!=!8!students!like!only!cartoons.!Adding!the!number!!of!students!in!these!3!categories!accounts!for!4!+!12!+!8!=!24!students.!Since!the!class!has!27!students,!the!number!of!students!who!do!not!like!either!is!27!–!24!=!3.!! METHOD 2: Strategy: Make a table to represent each student. Let!V!represent!the!students!who!like!video!games!and!let!C!represent!those!who!like!cartoons.! V!V!V!V!V!V!V!V!V!V!V!V!V!V!V!V!!!!!!!!!!!!!!!!C!C!C!C!C!C!C!C!C!C!C!C!C!C!C!C!C!C!C!C!!!!The!three!empty!spaces!at!the!end!of!the!table!represent!the!students!who!like!neither.! FOLLOW-UP: A survey of 91 fifth grade students found that 44 students like only dogs and 29 like only cats. If there are twice as many students who like both dogs and cats, as there are students who like neither, how many students like dogs? [56] 1D Strategy: Consider the rectangles without any overlap. If!the!two!rectangles!did!not!overlap,!the!total!length!would!be!13!+!13!=!26!cm.!If!we!slide!one!of!the!rectangles!over!the!other!until!the!total!length!is!22!cm,!there!would!be!26!–!22!=!4!cm!of!overlap.!Since!the!height!is!still!9!cm,!the!area!of!the!overlap!is!4!×!9!=!36$sq$cm.! 1E METHOD 1: Strategy: Use the shapes as items placed on a balance scale. Place!2!Δs!and!1!◊!on!one!side!of!the!scale!and!an!18!unit!weight!on!the!other!side.!Then!add!the!remaining!shapes!to!the!first!side!and!the!corresponding!weights!to!the!other!side!of!the!scale.!On!one!side!we!have!(Δ!+!Δ!+!◊)!+!(◊!+!◊!+!⌂) + (⌂!+!⌂!+!Δ)!and!on!the!other!side!18!+!22!+!17.!Therefore!three!of!each!shape!equals!57,!so!one!of!each!shape!equals!57/3!=!19.! METHOD 2: Strategy: Make a table. Δ!0!1!2!3!4!5$6!7!8!9!◊!18!16!14!12!10!8$6!4!2!0!⌂!X!X!X!X!2!6$10!14!18!22!Fill!in!the!table!with!all!possible!values!of!Δ!and!compute!the!other!values!based!upon!the!first!and!second!equations.!There!is!only!one!set!of!values!that!satisfy!the!third!equation.!These!are!shown!in!bold!and!5!+!8!+!6!=!19.! FOLLOW-UP: Given the equation 2 × Δ – 2 × ◊ + ⌂ = Δ – 2 × ◊ + 2 × ⌂ – 6, if Δ = 15, what is the value of ⌂? [21] NOTE: Other FOLLOW-UP problems related to some of the above can be found in our three contest problem books and in “Creative Problem Solving in School Mathematics.” Visit www.moems.org for details and to order. Olympiad 1, Continued

Division Contest for Elementary & Middle Schools Mathematical Olympiads December 16, 2014 E 2 Copyright © 201 4 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved. 2A METHOD 1 : Strategy: Pair numbers that add to 100. 9 + 26 + 83 + 55 + 45 + 17 + 74 + 91 = (9 + 91) + (26 + 74) + (83 + 17) + (55 + 45) = 4(100) = 400. METHOD 2: Strategy : Add the numbers in the order written. 9 + 26 + 83 + 55 + 45 + 17 + 74 + 91 = 400. FOLLOW -U P: Find the value of . [ 20 ] 2 B METHOD 1 : Strategy: Consider which numbers have an odd number of factors. The number 6 factors into 6 × 1 or 3 × 2. Therefore 6 has four factors, 1, 2, 3, and 6. The number 9 factors into 9 × 1 and 3 × 3. The number 9 has only three factors, 1, 3, and 9. To have an odd number of factors, the number must factor into two factors that are the same. Therefore it must be a perfect square. Count the number of perfect squares less than 20. These numbers are: 1, 4, 9, and 16. Therefore there are 4 numbers less than 20 that have an odd number of factors. METHOD 2: Strategy : List the factors of all non -prime numbers less than 20. Prime numbers only have 2 factors. 1: 1 9: 1, 3, 9 15: 1, 3, 5, 15 4: 1, 2, 4 10: 1, 2, 5, 10 16: 1, 2, 4, 8, 16 6: 1, 2, 3, 6 12: 1, 2, 3, 4, 6, 12 18: 1, 2, 3, 6, 9, 18 8: 1, 2, 4, 8 14: 1, 2, 7, 14 The four bolded results each have an odd number of factors. FOLLOW -U P: How many positive integers less than 50 have exactly two factors? [ 1 5] 2 C Strategy : Determine the least common denominator for the two fractions. The least common denominator for the fractions 2/3 and 3/4 is 12. Therefore the fewest possible number of games played in each half is 12. Team Alpha won games in the first half and games in the second half. Thus the fewest number of games won for the season was 8 + 9 = 17 games. 400 4 17 66 sq cm 22 SOLUTIONS AND ANSWERS 2A 2B 2C 2D 2E

2D METHOD 1 : Strategy : Label the cubes. There are 4 cubes with 5 red faces, 10 cubes with 4 red faces, and 2 cubes with 3 red faces. The total area of all the red faces is 4 × 5 + 10 × 4 + 2 × 3 = 20 + 40 + 6 = 66 sq cm. METHOD 2: Strategy : Divide the object into similar shaped pieces. The 4 prongs of three cubes each have 4 × 13 = 52 painted faces. The cubes that join these sets of three cubes each have 3 painted faces and 2 × 3 = 6. The two cubes joining the two groups of seven cubes each have 4 faces painted red and 2 × 4 = 8. The total area of all the painted cubes is 52 + 6 + 8 = 66 sq cm. METHOD 3: Strategy : Subtract the number of unpainted faces from the total number of faces. Sixteen cubes have a total of 16 × 6 = 96 sq cm. There are 15 places where one cube shares a face with another cube. Since these faces are common to two cubes, we subtract 2 × 15 = 30 from 96 to get 96 – 30 = 66 sq cm of painted cubes. 2E Strategy : Start with cases that factor uniquely. The only way to get 14 with three different factors is 14 = 1 × 2 × 7. The only way to get 15 with three different factors is 15 = 1 × 3 × 5. Therefore the upper left corner must be 1. Since 5 must be in the top r ow, it must be in the third column because 5 is a factor of 180 but not of 144. In column 1, either the 2 or the 7 must be in the third row. The 7 must be in the lower left corner since 7 is not a factor of 64. Note that 378/7 = 54 so the remaining numbe rs in the third row are 6 and 9. If 6 were in the lower right corner the number above it would be 180/(5 × 6) = 6, which is not possible. The fourth corner is 9 and the number above it will be 180/(5 × 9) = 4. The sum of the numbers in the 4 corners is 1 + 5 + 7 + 9 = 22 . NOTE: Other F OLLOW -UP problems related to some of the above can be found in our three contest problem books and in “Creative Problem Solving in School Mathematics.” Visit www.moems.org for details and to order. Olympiad 2, Continued

for Elementary & Middle Schools Mathematical Olympiads January 13, 2015 Copyright © 2014 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved. 3A METHOD 1: Strategy: Regroup and then add. Regroup!as!follows:!100!+!(91!–!82)!+!(73!–!64)!+!(55!–!46)!+!(37!–!28)!+!(19!–!10).!The!required!sum!becomes!100!+!5!×!9!=!145.$ METHOD 2: Strategy: Add and subtract from left to right. 100!+!91!–!82!+!73!–!64!+!55!–!46!+!37!–!28!+!19!–!10!=!145.! FOLLOW-UP: Use the algebraic fact that a 2 − b 2 = a + b ( ) a − b ( ) to compute the simplified value of 100 – 81 + 64 – 49 + 36 – 25 + 16 – 9 + 4 – 1. [55] 3B METHOD 1: Strategy: Recall that the word “of” in this context means multiplication. Since!“of”!means!multiplication!when!used!with!fractions,!cancel!first!and!then!multiply!the!remaining!terms:!.! METHOD 2: Strategy: Work backwards and use a table. 1.!Find!4/6!of!3000.!!!!2.!Find!3/5!of!2000.!!3.!Find!2/4!of!1200.!!4.!Find!1/3!of!600.!!The!final!answer!is!200.!!! FOLLOW-UP: Compute the integer value of of of of of of 1024 × 128. [4] 3C METHOD 1: Strategy: Divide the shape into smaller more familiar shapes. Separate!the!“H”!into!two!10!×!2!rectangles!and!one!2!×!2!square.!The!sum!of!the!areas!is!2(20)!+!4!=!44!square!units.!Since!each!box!contains!6!square!units!and!,!the!fewest!number!of!boxes!needed!to!tile!the!“H”!is!8.! METHOD 2: Strategy: Subtract the area of the regions from the rectangle. Find!the!area!of!the!6!×!10!rectangle!and!subtract!from!it!the!areas!of!the!two!2!×!4!rectangles.!The!total!area!of!the!“H”!is!6!×!10!–!2(2!×!4)!=!60!–!16!=!44!square!units.!Find!the!number!of!required!boxes!of!tiles!using!the!division!in!Method!1.! FOLLOW-UP: Find the number of units in the perimeter of the “H”. [48] 145 200 8 13 6 ! 1 3 × 2 4 × 3 5 × 4 6 × 3 0 0 0 = 1 1 5 × 3 0 0 0 = 2 0 0 ! 1 2 ! 1 4 ! 1 8 ! 1 1 6 ! 1 3 2 ! 4 4 6 = 7 2 6 SOLUTIONS AND ANSWERS 3A 3B 3C 3D 3E 3! 500 500 500 500 500 500 400 400 400 400 400 200 200 200 300 300 300 300 200 1 1 1 1 1 200$

3D METHOD 1: Strategy: Work backwards and use a table. !The!number!of!strawberries!must!be!a!multiple!of!4!and!also!3!more!than!a!multiple!of!5.!!! !!The!number!of!girls!on!the!team!is!13.! METHOD 2: Strategy: Use algebra. Let!G!be!the!number!of!girls!on!the!team.!Then!5G!+!3!=!4(G!+!4).!Apply!the!distributive!property!to!get!5G!+!3!=!4G!+!16,!so!G!=!13.! FOLLOW-UPS: (1) How many strawberries are there? [68] (2) Find the least integer that leaves a remainder of 1 upon division by each of 2, 3, 4, and 5. [61] 3E Strategy: Apply the divisibility rules. A!number!is!divisible!by!36!when!it!is!divisible!by!both!4!and!9.!!A!number!is!divisible!by!4!when!the!number!formed!by!the!last!two!digits!is!a!multiple!of!4.!!Since!0B!is!the!number!formed!by!the!last!2!digits,!B!must!be!0!or!4!or!8.!!A!number!is!divisible!by!9!when!the!sum!of!all!its!digits!is!a!multiple!of!9.!!The!sum!of!the!known!digits!is!3!+!8!+!7!+!2!+!6!+!4!+!9!+!0!=!39.!The!sum!of!the!remaining!two!digits!A!and!B!must!either!be!6!to!create!a!total!of!45!or!15!to!result!in!a!sum!of!54.!Any!sum!greater!than!15!cannot!be!the!result!of!adding!two!single!digit!numbers.!If!B!=!0,!then!A!=!6!to!have!A!+!B!=!6.!The!sum!of!15!would!not!be!possible.!If!B!=!4,!then!A!=!2!to!have!A!+!B!=!6.!The!sum!of!15!would!not!be!possible.!If!B!=!8,!then!A!=!7!to!have!A!+!B!=!15.!The!sum!of!6!would!not!be!possible.!This!last!case!violates!the!given!information!that!A!and!B!are!both!even.!Therefore!the!sum!A!+!B!=!6.!NOTE: Other FOLLOW-UP problems related to some of the above can be found in our three contest problem books and in “Creative Problem Solving in School Mathematics.” Visit www.moems.org for details and to order. Olympiad 3, Continued #!on!Team!#!of!strawberries!Team!+!4!coaches!#!of!strawberries!Check!17!17!×!5!+!3!=!88!21!21!×!4!=!84!88!≠!84!16!16!×!5!+!3!=!83!20!20!×!4!=!80!83!≠!80!15!15!×!5!+!3!=!78!19!19!×!4!=!76!78!≠!76!14!14!×!5!+!3!=!73!18!18!×!4!=!72!73!≠!72!13!13!×!5!+!3!=!68!17!17!×!4!=!68!68!=!68!

for Elementary & Middle Schools Mathematical Olympiads February 10, 2015 Copyright © 2014 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved. 4A METHOD 1: Strategy: Group nickels and dimes together. Pair!up!every!nickel!with!a!dime.!There!are!five!dimes!left!over,!and!those!are!worth!$.50.!Therefore!the!pairs!of!nickels!and!dimes!make!up!a!total!of!$1.40!–!$.50!=!$.90.!Each!nickelNdime!pair!is!worth!$.15,!so!there!are!90/15!=!6!pairs!of!nickels!and!dimes.!Pablo!has!6!nickels.! METHOD 2: Strategy: Use an organized guess and check. Since!the!sum!of!all!the!coins!does!not!end!in!a!five,!there!must!be!an!even!number!of!nickels.!If!there!are!2!nickels!and!7!dimes!the!total!amount!is!$.80.!With!4!nickels!and!9!dimes,!the!total!is!$1.10.!With!6!nickels!and!11!dimes!the!total!amount!is!$1.40.! FOLLOW-UP: On a plane are 150 passengers. There are 20 more men than women. How many women are on the plane? [65] 4B METHOD 1: Strategy: Apply the distributive property. Recognize!that!(18!×!40)!+!(12!×!40)!=!(18!+!12)!×!40!=!30!×!40.!Then!recognize!that!!(20!×!28)!+!(20!×!12)!=!20!×!(28!+!12)!=!20!×!40.!Add!30!×!40!+!20!×!40!=!(30!+!20)!×!40!=!50!×!40!=!2000.!$ METHOD 2: Strategy: Create a common factor. (18!×!40)!+!(12!×!40)!+!(20!×!28)!+!(20!×!12)!=!!(18!×!40)!+!(12!×!40)!+!(40!×!14)!+!(40!!×!!6)!=!40(18!+!12!+!14!+!6)!=!40(50)!=!2000.! FOLLOW-UP: What is the value of (15 × 20) + (47 × 20) + (38 × 20) + (100 × 12) + (100 × 5) + (100 × 3)? [4000] 4C METHOD 1: Strategy: Use the least common multiple (LCM). If!a!number!is!divisible!by!3,!5,!and!7,!it!is!divisible!by!their!least!common!multiple.!The!LCM!of!3,!5,!and!7!is!their!product,!since!the!three!numbers!have!no!common!factors!other!than!1.!Since!3!×!5!×!7!=!105,!which!is!not!between!200!and!300,!we!must!multiply!this!result!by!2.!This!gives!us!105!×!2!=!210!which!is!divisible!by!3,!5,!and!7!and!is!also!between!200!and!300.!The!number!is!210.! METHOD 2: Strategy: Consider the multiples of 7 between 200 and 300. We!first!consider!multiples!of!7!to!lessen!the!number!of!possibilities.!The!multiples!of!7!in!the!given!interval!are!203,!210,!217,!…,!294.!Since!the!number!must!be!divisible!by!5!as!well!it!must!end!in!either!a!5!or!a!0.!A!number!that!is!divisible!by!3!has!the!sum!of!its!digits!divisible!by!3.!The!number!210!satisfies!all!three!conditions!so!210!is!the!answer. 6 2000 210 10 181 sq cm SOLUTIONS AND ANSWERS 4A 4B 4C 4D 4E

4D Strategy: Count in an organized fashion. If!the!first!digit!is!6,!the!other!two!digits!must!add!to!18.!There!is!only!one!such!3Ndigit!number!whose!digit!sum!is!24.!That!number!is!699.!If!the!first!digit!is!7,!the!other!two!digits!must!add!to!17.!There!are!two!such!numbers:!789!and!798.!If!the!first!digit!is!8,!the!remaining!two!digits!must!add!to!16.!We!can!use!7!and!9!or!8!and!8.!There!are!three!such!numbers:!879,!897,!and!888.!If!the!first!digit!is!9,!the!other!two!digits!add!to!15.!There!are!four!such!numbers:!969,!996,!987,!and!978.!The!total!number!of!counting!numbers!for!which!the!digit!sum!is!24!is!1!+!2!+!3!+!4!=!10.! 4E METHOD 1: Strategy: Count the painted areas for corner and edge cubes separately. The!top!cube!has!5!painted!faces,!a!total!of!5!sq!cm!on!that!layer.!!For!each!layer!below!the!top,!there!are!two!kinds!of!cubes:!corner!cubes!!that!have!3!painted!faces!and!edge!cubes!which!have!only!2!painted!!faces.!Each!layer!will!have!4!corner!cubes!so!there!are!4!×!4!=!16!corner!!cubes!for!a!total!of!16!×!3!=!48!sq!cm.!!On!a!layer!with!n!cubes!on!a!side,!there!will!be!n!–!2!edge!cubes!on!each!of!four!sides.!Therefore!in!the!3!×!3!layer!there!are!4!×!(3!–!2)!=!4!edge!cubes!for!a!total!of!4!×!2!=!8!sq!cm.!In!the!5!×!5!layer!there!is!a!total!of!4!×!(5!–!2)!=!12!edge!cubes!with!an!area!of!12!×!2!=!24!sq!cm.!The!remaining!two!layers!have!areas!of!4!×!(7!–!2)!×!2!=!40!sq!cm!and!4!×!(9!–!2)!×!2!=!56!sq!cm.!The!total!painted!area!is!5!+!48!+!8!+!24!+!40!+!56!=!181$sq$cm."! METHOD 2: Strategy: Look at the picture from different angles. Looking!down!from!the!top,!the!exposed!area!is!just!the!area!of!a!9!×!9!!rectangle!or!81!sq!cm.!Look!at!the!shape!from!the!side!and!the!exposed!!area!is!1!+!3!+!5!+!7!+!9!=!25!sq!cm.!Since!there!are!4!sides!the!total!area!that!would!be!painted!is!81!+!4!×!25!=!181!sq!cm.! FOLLOW-UP: Continue the pyramid by adding rows beneath the 9 by 9 row. Follow the same pattern for 5 more rows. Find the volume of the new pyramid. [1330 cu cm] NOTE: Other FOLLOW-UP problems related to some of the above can be found in our three contest problem books and in “Creative Problem Solving in School Mathematics.” Visit www.moems.org for details and to order. Olympiad 4, Continued

for Elementary & Middle Schools Mathematical Olympiads March 3, 2015 Copyright © 2014 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved. 5A METHOD 1: Strategy: Re-write the problem. Decompose!the!addends!using!base!ten.!100!+!200!+!300!+!400!+!500!+!600!+!700!+!800!+!900!=!4500!2!+!3!+!4!+!5!+!6!+!7!+!8!+!9!+!1!=!45!Now!add!4500!+!45!to!get!4545.! METHOD 2: Strategy: Form pairs of equal values. Note!that!the!sum!of!102!+!809!=!911!and!203!+!708!=!911.!There!are!four!such!pairs!of!911.!Therefore!the!sum!of!the!eight!addends!is!4!×!911!=!3644.!Now!add!the!last!addend!to!get!3644!+!901!=!4545.! FOLLOW-UP: Combine: 908 – 807 + 706 – 605 + 504 – 403 + 302 – 201. [404] 5B Strategy: Find the pattern. The!first!number!times!the!second!number!plus!the!third!number!equals!the!number!in!the!bottom!row.!Therefore,!7!×!2!+!3!=!14!+!3!=!17.!!! 5C Strategy: Make a table. The!value!of!the!number!of!quarters!has!to!be!a!multiple!of!$.25.!The!number!of!quarters!has!to!be!odd!since!the!total!ends!in!a!5.!Since!the!number!of!quarters!is!greater!than!the!number!of!dimes,!start!the!table!with!13!quarters.!!A!(Number!of!quarters)!13!11!9!B!(Number!of!dimes)!2!7!12!Total!amount!$3.45!$3.45!$3.45!!Since!the!number!of!quarters!(9)!is!not!more!than!the!number!of!dimes!(12)!in!the!last!column,!there!are!only!2!possible!values!for!A.! FOLLOW-UP: Kesha has some nickels and dimes. She has a total of 19 coins with more nickels than dimes. What is the greatest amount of money Kesha can have? [$1.40] 4545 17 2 2 1/2 100 sq cm SOLUTIONS AND ANSWERS 5A 5B 5C 5D 5E

Powered by TCPDF (www.tcpdf.org) 5D METHOD 1: Strategy: Draw a diagram. Draw!a!rectangle!and!divide!it!into!thirds!and!fifths.!Shade!in!1/5!or!3/15!!of!the!rectangle!in!dark!grey.!This!is!the!amount!that!the!small!hose!will!fill!!in!the!first!hour!when!running!alone.!In!the!next!hour,!the!small!hose!will!!again!fill!1/5!of!the!pool!(light!grey)!while!the!large!hose!fills!1/3!or!5/15!!of!the!pool!(diagonal!lines)!in!that!same!hour.!The!white!region!still!needs!to!be!filled!while!the!small!hose!and!the!large!hose!are!both!turned!on.!The!white!portion!is!4/15!of!the!pool.!The!light!grey!and!the!diagonal!lines!regions!represent!the!amount!of!the!pool!filled!by!both!devices!in!1!hour!and!is!8/15!of!the!pool.!Therefore!to!fill!4/15!of!the!pool!it!will!take!1/2!hour.!The!total!time!to!fill!the!pool!is!1!+!1!1/2!hours!or!2&1/2!hours!all!together.! METHOD 2: Strategy: Apply some algebra. Let!t!be!the!number!of!hours!the!small!hose!is!running.!!Then!t!–!1!is!the!number!of!hours!the!large!hose!is!running.!!In!t!hours!the!small!hose!fills!(1/5)!×!t!of!the!pool.!!In!t!–!1!hours,!the!large!hose!will!fill!(1/3)!×!(t!–!1).!Together!(1/5)!×!t!+!(1/3)!×!(t!–!1)!=!1!full!pool.!If!we!multiply!by!15!we!get!3t!+!5(t!–!1)!=!15!so!!8t!–!5!=!15,!8t!=!20,!and!t!=!20/8!=!2.5!hours.! 5E METHOD 1: Strategy: Subtract the area of the non-shaded regions from the total area. Since!the!area!of!the!square!is!180!square!centimeters!and!there!are!36!squares,!the!area!of!each!square!is!180/36!=!5!square!centimeters.!Count!the!number!of!unshaded!squares!and!triangles.!There!are!8!squares!and!16!triangles.!Since!two!triangles!can!form!one!square,!the!total!unshaded!area!is!8!×!5!+!8!×!5!=!80!square!centimeters.!Thus!the!shaded!area!is!180!–!80!=!100!sq!cm.![Note:!Due!to!symmetry!you!can!work!with!1/4!of!the!region!and!then!multiply!by!4.]! METHOD 2: Strategy: Count the shaded areas. Find!the!area!of!each!square!as!in!Method!1.!Then!count!the!number!of!squares!and!triangles!in!the!shaded!region.!There!are!12!squares!and!16!triangles.!Since!two!triangles!can!be!used!to!form!a!square,!the!total!area!is!12!×!5!+!8!×!5!=!60!+!40!=!100!square!centimeters.! FOLLOW-UP: An elementary school has a rectangular courtyard with a 3-foot wide path going around it. The length of the courtyard is 20 feet and the width is 10 feet. The school wants to tile the entire path. Each tile is 3-feet by 1-foot. How many tiles will they need? [72] NOTE: Other FOLLOW-UP problems related to some of the above can be found in our three contest problem books and in “Creative Problem Solving in School Mathematics.” Visit www.moems.org for details and to order. Olympiad 5, Continued