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© Copyright 2016, MOEMS® . All rights reserved. 201ò TOURNAMENT Grade 6 and Below Table of Contents Individual Event: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Individual Event: Student Answer Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Team Event: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Team Event: Student Answer Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Scoring Room Key . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Individual Event: Model Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6-7 Team Event: Model Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8-9 Tiebreakers 1-5: Student Sheets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10-14 Tiebreakers: Problems and Model Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 PRACTICE PROBLEMS FOR 201 9

© Copyright 2016, MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. Division 456 - 1 2016 Individual Event No calculators are permitted during this tournament.Time Limit: 30 minutes. 1. The sum of two consecutive multiples of 7 is 49. What is the greater of these numbers? 2. Terry walks 4/5 of the way from school to home in exactly 20 minutes. How long does it take for her to walk the rest of the way home? 3. A plane flies at 660 feet per second (ft/sec). How many miles per hour (mph) is this? Note: 88 ft/sec is the same as 60 mph. 4. A number of identical wooden blocks are piled on each other to form a cube that is 12 centimet\ ers on each side. Each block is a rectangular solid that is 2 centimeters by 3 centimeters by 6 centimeters. How many blocks are there? 5. One sandwich and a drink costs $7.95. At the same prices, three sandwiches and a drink cost $19.95. What is the price of a drink? 6. For every $6 Emily has, Megan has $4 and Grace has $3. Olivia has the re\ st of the money. If the four girls have a total of $61 in dollar bills only, what is the least number of dollar bills Olivia can have? 7. ABCD represents a four-digit number. For how many different odd numbers is A an even digit if no two digits are the same? 8. The total surface area of a cube is 96 square centimeters. The cube is enlar ged so that each edge is increased by 6 centimeters. By how much is the total surface area of the\ resulting cube increased? 9. This addition at the right shows three 3-digit numbers. Every digit from 1 through 9 is used exactly one time. Each letter represents a digit. What 3-digit number is represented by the sum BD4? 10. Of the 45 people in the school band, 17 do not ride to school and 22 are\ boys. If 15 of the girls ride to school, how many boys in the band do not ride to school? 2 C E + A 3 F B D 4

© Copyright 2016, MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. Division 456 - 2 SCORE Do not write in this space. Student Name School and Team Team # 2016 Individual Event: Answers Write answers clearly. Each correct answer will receive one point. 1. 2. 3. 4. 5. 6. dollar bills 7. 8. 9. 10. minutes $ mph blocks numbers sq cm boys

© Copyright 2016, MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. Division 456 - 3 2016 Team Event No calculators are permitted during this tournament. Time Limit: 20 minutes. 11. What is the tens digit in the product of 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 ×\ 11 × 12? 12. I am 17 years younger than my aunt. If her age is 5 years less than three times mine, how old\ am I? 13. I have only coins but I cannot make change for a dollar, a half-dollar, a quarter , a dime, or a nickel. What is the greatest possible value of my coins? (Assume the half-dollar is the greatest coin in value. ) 14. The perimeter of a rectangle is 44 meters. The length is 4 meters greater than the width. What is the area of the rectangle? 15. If Ana’s age is added to the square of Billy’ s age, the sum is 53. But if Billy’s age is added to the square of Ana’s age, the sum is 23. What is the sum of Ana’s and Billy’s ages? 16. A store discounts by always marking down each item by the same percent. The original price of one item was $32. The first discount resulted in a price of $24. The second discount resulted in a price of $18. What price resulted from the third discount? 17 . Taylor has $5.00, all in dimes, quarters, or a combination of dimes and quarters. Which of the following cannot be the total number of coins that she has? 23, 29, 35, 37, 41, 50 18. A rectangle, a circle, and a regular pentagon are placed so that they overlap. No side of the rectangle coincides with a side of the pentagon. What is the greatest possible number of points of intersection? 19. What is the least counting number that is greater than 200 and has exactly three different factors? 20. The length, width, and semi-perimeter of a rectangle are each a prime number. The length is 13 cm more than eight times the width. What is the area of the rectangle? (The semi-perimeter is one-half of the perimeter.)

© Copyright 2016, MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. Division 456 - 4 School and Team Team # Student Names 2016 Team Event: Answers Write answers clearly. Each correct answer will receive one point. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. SCORE Do not write in this space. years $ sq m $ sq cmyears coins points

SCORING ROOM KEY 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. © Copyright 2016, MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. Division 456 - 5 2016 0 11 1.19 117 11 13.50 37 26 289 58 28 5 450 48 1.95 9 or 9.00 1120504 954 9 INDIVIDUAL Event: Answers TEAM Event: Answers

© Copyright 2016, MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. Division 456 - 6 2 2 2 2 2 2 3 3 3 3 6 6 INDIVIDUAL EVENT SOLUTIONS, 2016 ANSWERS: 1) 28 2) 5 3) 450 4) 48 5) 1.95 6) 9 or 9.00 7) 1 120 8) 504 9) 954 10) 9 1. METHOD 1: 7KH ?UVW IHZ PXOWLSOHV RI7DUH 0714 21 28 35 42 49 7KH WZR PXOWLSOHV RQ WKH OLVW WKDW  DUH FRQVHFXWLYH DQG DGG XS WR49 DUH 21 DQG 28 7KH JUHDWHU RIWKHVH QXPEHUV LV28 . METHOD 2: +DOI RI49 LVEHWZHHQ 24 DQG 25 7KH QHDUHVW WZR PXOWLSOHV RI7DUH 21 DQG 28 7KH\ DUH  FRQVHFXWLYH DQG WKHLU VXP LV49 7KH JUHDWHU PXOWLSOH RI7LV28 2. METHOD 1:  7HUU\ KDG ZDONHG O PRIWKH ZD\ KRPH 6KH KDV DQRWKHU L PRIWKH ZD\ OHIW %HFDXVH one?IWK  LV  L ORIfour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x 88 60     88[  39600      [ 450 PSK 4. METH OD 1: 7KH YROXP H RIWKH 12 E\ 12 E\ 12 FXE H LV1728 FX ELF FH QWLPHWHUV  FF  7KH YROXPH RIRQH 2E\ 3E\ 6EORFN LV36 FF 7KHQ WKHUH DUH 1728 ?36   48 48 EORFNV LQWKH FXEH

© Copyright 2016, MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. Division 456 - 7 METHOD 2: Suppose every block in the cube is placed in the same direction as shown. The front of the cube shows 2 identical stacks that are 6 blocks high. That is 12 blocks. But the cube is four such sets of blocks deep. Therefore, there is a total of 12 × 4 = 48 blocks in all. 5. If two sandwiches are added to a $7.95 order, they result in a $19.95 order. This costs $12 more, so each sandwich costs $6. Therefore, one drink costs $7.95 – $6.00 = $ 1.95 . 6. For every $6 Emily has, Emily , Megan and Grace have a total of $13. Thus the most they can have is 4 × 13 = $52 and the least Olivia can have is 61 – 52 = $ 9 . 7. ABCD is an odd number, so D can be any of 5 digits: 1, 3, 5, 7, or 9. Since A is even and ABCD has four digits, A can be any of four digits: 2, 4, 6, or 8, but not 0. This leaves any of eight digits for B and any of seven for C. Thus, 4 × 8 × 7 × 5 = 1120 different four-digit numbers are possible. 8. There are six congruent surfaces to a cube. Before the cube was enlarged, the area of one surface was 96 ÷ 6 = 16 sq cm. Each surface is a square so the length of one edge was 4 cm. After the cube is enlarged, the length of one edge is 10 cm. Then the area of one surface is 100 sq cm and the total surface area is 600 sq cm. The increase is 600 – 96 = 504 sq cm. 9. The missing digits are 1, 5, 6, 7, 8, and 9. Thus, E + F can only be 5 + 9 or 6 + 8 in some order . Either way, the tens column is increased by 1. Thus C + 4 ends in D; so C,D in that order is 1,5 or 5,9 or 7,1. However, the 5 can only be used once. If C is 5 or 7, then the hundreds column is increased by 1 and 3 + A = B. Thus A,B = 5,8 or 6,9. But if C is 1, the hundreds column is not increased and 2 + A = B, so that A,B is 5,7 or 6,8 or 7,9. Of all these choices, only 216 + 738 = 954 or 218 + 736 = 954 uses all nine of the digits, each once. Either way, the sum is 954 . 10. METHOD 1: Because 15 girls do ride to school, consider two overlapping groups: the girls and the riders. There are 45 – 22 = 23 girls and 45 – 17 = 28 riders, with 15 in b\ oth groups. How many belong to only one group? 23 – 15 = 8 girls are not riders and 28 – 15 = 13 riders are not girls (they are boys). Since 13 of the 22 boys ride to school, 9 boys do not ride to school. METHOD 2: As in me thod 1, there are 23 girls and 28 riders, with 15 people being counted twice since they are in both groups. This accounts for a total of 23 + 28 – 15 = 36 people. The remaining 45 – 36 = 9 people are neither girls nor riders, so there are 9 boys who do not ride to school. 23 girls 28 riders 15 8 13 9 boy non-riders 2 C E + A 3 F B D 4

© Copyright 2016, MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. Division 456 - 8 TEAM EVENT SOLUTIONS, 2016 ANSWERS: 1) 0 2) 11 3) 1.19 4) 11 7 5) 11 6) 13.50 7) 37 8) 26 9) 289 10) 58 11. Since 5 × 6 = 30 and 30 × 10 = 300, the final product is a multiple of 300. The tens digit is 0 . 12. METHOD 1: My aunt is more than 17 years old. So I must be at least (18 + 5) ÷3 = almost 8 years old. In the table below, the first row lists my possible ages, the second row triples my age and then subtracts 5 years and the third row lists the difference in years between our ages. My age 8 9 10 11 My aunt’s age 24 – 5 = 19 22 25 28 Difference 1113 15 17 Only when I am 11 years old is she 17 years older than me. METHOD 2: Let x represent my age in years. Then 3x – 5 represents my aunt’s age. Equation: Her age minus my age = 17 years: (3x – 5) – x= 17 Combine like terms: 2x – 5 = 17 Add 5 to each side of the equation: 2x = 22 Divide both sides of the equation by 2: 2x=11 I am 11 years old. 13. Start with the least coin and add the others one at a time. I cannot have more than 4 pennies or else I could make change for a nickel. If I then also use four dimes, one quarter and one half-dollar, I would not be able to make change for any of the amounts listed. The greatest possible value is $1.19 . 14. The semi-perimeter is half the perimeter because it is the sum of one length and one width. Then the semi- perimeter is 22 m. Since the length is 4 m greater than the width, we need two numbers that add to 22 and subtract to 4. The rectangle is 13 m long and 9 m wide. The area is 117 sq m. 15. The square of Billy’s age is less than 53, namely 49, 36, 25, 16, 9, 4, or 1. Thus, Billy is 7 years old or less. If Billy were 7, Ana would be 53 – 49 = 4. Similarly, if Billy were 6, 5, or 4, Ana would be 53 – 36 = 17, 53 – 25 = 28, or 53 – 16 = 37. On the other hand, the square of Ana’s age is less than 23. Of 4, 17, 28, and 37, only the square of 4 is less than 23. The sum of Ana’s and Billy’s ages is 4 + 7 = 11 years. 16. After applying the first discount, $24 is 24 32 of $32. The second price is 3 4 or 75% of the first price. (The discount itself is 25% or 1 4.) Applying the second discount, 3 4 of $24 is $18, as given. Then applying the third discount, 3 4 of $18 is $13.50 .

© Copyright 2016, MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. Division 456 - 9 17. Start with $5 all in quarters, trade quarters for dimes and examine the total numbers of coins. If Taylor has 20 quarters, she would have 0 dimes, for a total of 20 coins. Since two quarters have the same value as five dimes, Taylor can have 20, 18, 16, …, 4, 2, 0 quarters. Consequently , she would have 0, 5, 10, …, 50 dimes respectively. Then she would have a total of 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50 coins. Thus, she cannot have 37 coins. 18. Draw the figures in pairs, as shown. Make sure that all three figures do not pass through any one point. In each case trace one figure and think of it as entering and then leaving the other figure; this means there must be an even number of points of intersection. The circle leaves and then reenters each of the four sides of the rectangle, so they have 8 points in common. The circle leaves and then reenters each of the five sides of the pentagon, so they have 10 points in common. The pentagon leaves and then reenters each of the four sides of the rectangle, so they have 8 points in comm on (rota ting the rect angle would not add two more point s).Thus, the great est possibl e number of points of intersection is 8 + 10 + 8 = 26 . 19. Examine simpler numbers: 1, 2, 3, 4, … The number 4 has just three factors: 1 and 4 itself and 2 because 2 × 2 = 4. Similarly, the number 9 has three factors: 1 and 9 itself and 3 because 3 × 3 \ = 9. So two factors are 1 and the number itself, and the third factor is the square root of our number. Thus, we are looking for a perfect square number. But 16 has five factors: 1, 2, 4, 8, and 16, because 4, the square root of 16, is not a prime number. Check the squares of prime numbers: the factors of 25 are 1 and 25, and 5. Also, the factors of 49 are 1 and 49, and 7. The following table lists the squares of all the prime numbers less than\ 300. Prime number 2 3 5 7 1113 17 Square of the prime 4 9 25 49 121 169 289 Factors 1, 2, 4 1, 3, 9 1, 5, 25 1, 7, 49 1, 11, 121 1, 13, 169 1, 17, 289 The least number which is greater than 200 and has just three factors is\ 289 . 20. Every prime number except 2 is odd. An odd number is the sum of an odd number and an even number. Thus, the width is 2 cm, and both the length and the semi-perimeter are consecutive odd prime numbers, since the semi-perimeter is the sum of one length and one width. Then the length is 8 × 2 + 13 = 29 cm and the semi- perimeter is 2 + 29 = 31 cm; both 29 and 31 are prime numbers. The area of the rectangle is 2 × 29 = 58 sq cm.

© Copyright 2016, MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. Divisiion 456 - 10 Three different members of the set {4, 5, 6, 7, 8, 9} are chosen at random and added. How many different sums are possible? 1 sums Time Limit: 5 minutes. No calculators are permitted. Name ________________________________ Team _________________________ Answer _____________________ TIEBREAKER #

© Copyright 2016, MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. Divisiion 456 - 11 Ten days from a Sunday is a Wednesday. What day of the week is 100 days from a Wednesday? 2 Time Limit: 5 minutes. No calculators are permitted. Name ________________________________ Team _________________________ Answer _____________________ TIEBREAKER #

© Copyright 2016, MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. Divisiion 456 - 12 B 5 6 5 C 9 + 7 4 D A 0 0 0 In the addition below, what four-digit number is represented by ABCD? 3 ABCD = Time Limit: 5 minutes. No calculators are permitted. Name ________________________________ Team _________________________ Answer _____________________ TIEBREAKER #

© Copyright 2016, MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. Divisiion 456 - 13 How many different whole numbers can be formed from the digits 9, 7, 5, and 3? 4 numbers Time Limit: 5 minutes. No calculators are permitted. Name ________________________________ Team _________________________ Answer _____________________ TIEBREAKER #

© Copyright 2016, MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. Divisiion 456 - 14 5 The number 5,472,B68 is a multiple of 9. What digit does B represent? Time Limit: 5 minutes. No calculators are permitted. Name ________________________________ Team _________________________ Answer _____________________ TIEBREAKER #

© Copyright 2016, MOEMS ® - Mathematical Olympiads for Elementary and Middle Schools. All rights reserved. Division 456 - 15 B 5 65 C 9 + 7 4 D A 0 0 0 2016 TIEBREAKERS Questions are given one at a time. Winning places are awarded in the order that correct answers are submitted. Incorrect answers result in elimination. No calculators are permitted during this tournament. Time limit: 5 minutes per question. T1. Three different members of the set {4, 5, 6, 7, 8, 9} are chosen at random and added. How many different sums are possible? T2. Ten days from a Sunday is a W ednesday. What day of the week is 100 days from a Wednesday? T3. In the addition at the right, what four -digit number is represented by ABCD? T4. How many different whole numbers can be formed from the digits 9, 7, 5, and 3? T5. The number 95,472,B68 is a multiple of 9. What digit does B represent? SOLUTIONS ANSWERS: T1) 10 T2) Friday T3) 2695 T4) 24 T5) 4 T1. The sum is at least 15 and at most 24, and can be any whole number from 15 to 24 inclusive. Hence, 10 sums are possible. T2. METHOD 1: Form a calendar and count the days starting at a Thursday . The 100th day is a Friday . METHOD 2: Divide 100 by 7. There are 14 weeks and then two extra days. The two extra days are Thursday and Friday. T3. In the ones column, 6 + 9 = 15. Then to get a sum that ends in 0, D must be a 5 and 2 will be carried. In the tens column, 2 + 5 + 4 = 11. For a sum that ends in 0, C must be a 9 and 2 will be carried. In the hundreds column, 2 + 5 + 7 = 14. For a sum that ends in 0, B must\ be a 6 and 2 will be carried. Then A will be a 2. The four-digit number is 2695 . T4. METHOD 1: Any of the 4 digits can occupy the thousands place. For each choice, any of the remaining 3 digits can occupy the hundreds place. For each of the 4 × 3 = 12 arrangements, either of the remaining 2 digits can occupy the tens place. For each of the 4 × 3 × 2 = 24 arrangements\ , the remaining digit must occupy the units place. There are a total of 24 four digit numbers that can be formed. METHOD 2: List them: 3579 5379 7359 9357 3597 5397 7395 9375 3759 5739 7539 9537 3795 5793 7593 9573 3957 5937 7935 9735 3975 5973 7953 9753 T5. The sum of the digits must be a multiple of 9: 5 + 4 + 7+ 2 + B + 6 + 8 = 32 + B. Then B = 4 . Note: It is faster to ‘cross out nines’ (5 + 4 and 7 + 2) since each adds up to 9. Then just test B + 6 + 8 for divisibility by 9.