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2018 AMC Middle Primary Solutions Solutions { Middle Primary Division 1. 4 + 4 = 8, hence (C). 2. A= 3 3 = 9, B= 2 4 = 8, C= 2 + 3 + 2 = 7, D= 4 + 4 = 8, E= 5 + 5 = 10, hence (E). 3. 6 tens is sixty, and 3 ones is three, so sixty-three, hence (A). 4. 1911 = 8 so 11 + 8 = 19, hence (B). 5. The diameter is 5 mm past 20 mm, making 25 mm, hence (D). 6. (Also UP1) 208 is between 205 and 210. Of these, the distance to 205 is 3 and the distance to 210 is 2. Therefore 210 is closer, hence (D). 7. (Also UP4, J4) The back of the necklace will look like the mirror image of the front of the necklace. So each letter will be mirrored, and the order of the letters will be reversed: KATE KA TE front back hence (A). 8. Parliament House tours leave at 8.30, 8.45, 9.00, 9.15, 9.30, 9.45, and so on. National Museum tours leave at 8.30, 8.50, 9.10, 9.30, 9.50, and so on. So tours leave at the same time at 8.30, 9.30, 10.30, and so on, which is every 60 minutes, hence (E). 9. The number of votes for sh, dogs and rabbits in total is 4 + 14 + 3 = 21. So there were 2921 = 8 votes for cats, hence (D). c Australian Mathematics Trust www.amt.edu.au 33

2018 AMC Middle Primary Solutions 10. Alternative 1 Since one-half is two quarters, we can convert (A){(E) into quarters. (A) is 4 quarters, (B) is 6 quarters, (C) is 5 quarters, (D) is 3 quarters and (E) is 6 quarters. Only (A) is 4 quarters, or a whole, hence (A). Alternative 2 The values in (A){(E) can be sketched in comparison to the whole unit on the left: 1 1 2 14 14 (A) 1 2 12 1 4 14 (B) 1 2 14 14 1 4 (C) 1 2 14 (D) 1 4 14 14 14 1 2 (E) Then only (A) is equal to one whole, hence (A). 11. (Also UP7) Alternative 1 If she had 60 books, she would ll 5 shelves with 12 books each. Since she has 2 fewer, her last shelf has 10 books, hence (D). Alternative 2 58 12 = 4 r10 so that she lls 4 shelves with 12 books, with 10 books on the 5th shelf, hence (D). 12. Alternative 1 The total number of cubes is 3 3 3 = 27. The number that are visible are 9 in the top layer, 5 in the middle layer and 5 in the bottom layer, for a total of 19. Therefore 27 19 = 8 cubes are not visible, hence (B). Alternative 2 The hidden cubes form a 2 2 2 cube behind the visible cubes. This has 2 2 2 = 8 small cubes, hence (B). c Australian Mathematics Trust www.amt.edu.au 34

2018 AMC Middle Primary Solutions 13. Alternative 1 Working backwards:  2nd oor, up 7 oors to 9th oor  9th oor, down 6 oors to 3rd oor  3rd oor, up 5 oors to 8th oor So Shelley must have started on the 8th oor, hence (E). Alternative 2 Shelley's total lift travel is 6 oors up and 12 oors down, a net travel of 6 oors down. Since she ended on the 2nd oor, she must have started on the 8th oor, hence (E). 14. (Also UP9) The points for Zac and Bill are lower than Pat's, so their calls cost less. Since the point for Bill is to the right, his call is longer and cheaper than Pat's, hence (B). 15. (Also UP12) There are 20 squares in the pattern. Since there are no overlaps, shape (C), with 6 squares, can't be the answer. Shape (E) can't be the answer, since the the leftmost and rightmost squares can't be covered by this pattern. Shapes (B) and (D) can be ruled out by trying to cover the area without overlap, as shown below. (B) (D)(A) This leaves only (A), which can be done as shown, hence (A). 16. For KAREN, 5 letters are needed. For WARREN, set aside the K, and another two letters are needed: W and R. For ANDREW, set aside one R, and other letter is needed: D. In all the smallest possible set of letters is A, D, E, K, N, R, R, W, hence (C). c Australian Mathematics Trust www.amt.edu.au 35

2018 AMC Middle Primary Solutions 17. Slicing each pizza into thirds, there are 12 slices. Each person gets 2 slices, so there are 6 people, hence (B). 18. (Also UP13) The clock needs to show after 11.30123456789101112 and b efore 12.30 123456789101112 . Of the clock faces (A){(E) shown, only (A) is between these, hence (A). 19. Alternative 1 The hidden faces on each of the 3 dice can be added: 2 + 3 + 6 = 11 1 + 3 + 4 + 5 = 13 2 + 4 + 5 + 6 = 17 41 hence (C). A lternative 2 On three standard dice, there are (1 + 2 + 3 + 4 + 5 + 6) 3 = 63 dots. In the diagram 22 dots are showing, so 63 22 = 41 dots are hidden, hence (C). 20. (Also UP17) Any triangle must have the numbers 1, 2 and 3, for a total of 6. As the rst diagram shows, the total is at least 14. Also, 14 is only possible with 1 on the left and the right. 6 61, 2 or 3 1, 2 or 3 661, 2 or 3 1, 2 or 3 2 33 2 11 11 Similarly, the second diagram shows that 14 is only possible with 1 in the top and bottom circles. Trying 1 in all four outer circles leads to the third diagram, with total 14, hence (C). c Australian Mathematics Trust www.amt.edu.au 36

2018 AMC Middle Primary Solutions 21. Alternative 1 15 medium eggs weigh the same as 18 small eggs, and also the same as 10 large eggs. So 5 large eggs weigh the same as 9 small eggs, hence (D). Alternative 2 Suppose the small egg weighs 50 g, so that six small eggs weigh 300 g. Then one medium egg weighs 60 g and six medium eggs weigh 360 g. Then one large egg weighs 90 g and ve large eggs weigh 450 g. This is the same as 9 small eggs, hence (D). Note: For any other weight of a small egg, the other weights would be di erent, but in the same proportions, leading to the same answer. 22. (Also UP19) From the rst row, = 24 4 = 6. Comparing the rst two rows, = 1 = 5. Comparing the rst two columns, = 2 = 3. Then + = 6 + 3 = 9, hence (B). 23. Alternative 1 The hexagon can be divided into pieces that can be rearranged as shown. 24 3 243 24 24 3 3 3 3 3 3 The total area is 2 24 + 4 3 = 60 square centimetres, hence (D). Alternative 2 The hexagon is made of two trapeziums, each of area 1 2  6 (6 + 4) = 30 cm 2 . The total area is then 2 30 = 60 cm 2 , hence (D). c Australian Mathematics Trust www.amt.edu.au 37

2018 AMC Middle Primary Solutions 24. (Also UP22) Anh gets 1 4 of the coins, which is 20 4 = 5 coins, leaving 15. Then Brenda gets 1 3 of these coins, which is 15 3 = 5 coins, leaving 10. Then Chen gets half of these coins, as does Dimitris, which is 5 coins each. So they all get the same number of coins, hence (E). 25. The edge of the coloured square showing in the nal pattern is two strips from the edge of the paper. So the 20 cm is made up of 8 cm plus 4 strip-widths. That is, the width of the folded-over strip is 1 4 of 12 cm, which is 3 cm, hence (D). 26. Alternative 1 The total of Ari, Billy and Charlie combined is 15 + 18 + 13 = 46. In this total, the three regions A, B and C are hit twice each. So the total from hitting A, B and C once each is 23. Since the scores from A and C add to 15, region B must score 8, and then Davy scores 16, hence (16). Alternative 2 Together Billy and Charlie score 18 + 13 = 31, from region A once, region B twice and region C once. Together Ari and Davy also hit region A once, region B twice and region C once, so their combined score is also 31. Since Ari scored 15, Davy must have scored 31 15 = 16, hence (16). 27. Let the number be N. Since Nis divisible by 3 and by 11, it is divisible by 33. That is, Nis one of the numbers 33; 66;99;132; 165 and so on. Also, Nis one more than a multiple of 14, so Nis odd. So we check the odd multiples of 33: N 33 99 165 231 297 363 429 495 561 627 . . . N 1 32 98 divis. by 14?  X We notice that 98 is a multiple of 14. However this is not a 3-digit number, so it is not the solution. We could keep checking, but we don't have to. The next multiple of 14 will occur 7 steps later, since 7 66 = 462 is a multiple of 14. N 33 99 165 231 297 363 429 495 561 627 . . . N 1 32 98 560 divis. by 14?  X X c Australian Mathematics Trust www.amt.edu.au 38

2018 AMC Middle Primary Solutions So 561 is a 3-digit solution. The next solution is 561 + 462 = 1023, which is not a 3-digit number, so that 561 is the only solution, hence (561). 28. Alternative 1 The staircase can be sliced into vertical slices, each with a triangular number of blocks. For the 12-step staircase, each vertical slice is a triangle with 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 78 blocks, and there are 12 such slices. Therefore the 12-step staircase has 12 78 = 936 blocks, hence (936). Alternative 2 Slice the staircase halfway up and turn the top part over as shown. 12 6 6 12 12 6 1 12 6 6 7 The result is a rectangular prism with 6 12 13 = 936 blocks, hence (936). 29. (Also UP26) In the algorithm, there must be a carry from the 100s column to the 1000s column. In the units column, a+ b+ c= 8 or a+ b+ c= 18, with 0 or 1 carried into the 10s column. If a+ b+ c= 8, then there is no carry from the 1s column, and then the 10s or 100s columns don't have a carry either. This makes the 1000s column add to 1, not 2. So this case is eliminated. So a+ b+ c= 18 and 1 is carried from the 1s column to the 10s column. In the 10s column, 1 + a+ b= 1 or 1 + a+ b= 11, so that a+ b= 0 or a+ b= 10. However, if a+ b= 0, then a= b= 0, which is too small to have a+ b+ c= 18. Thus a+ b= 10 and c= 8. In the 100s column, 1 + a= 10, so that a= 9. Then b= 1. Therefore the 3-digit number abcis 918, hence (918). c Australian Mathematics Trust www.amt.edu.au 39

2018 AMC Middle Primary Solutions 30. (Also UP28) From 1 to 99 there are 9 zeros, from 10; 20; : : : ;90. From 100 to 199 there are 20 zeros, ten being units digits and ten being tens digits. We can count these within the `century' of 100s. 100 # 101 # 102 # 103 # 104 # 105 # 106 # 107 # 108 # 109 # 110 " 120 " 130 " 140 " 150 " 160 " 170 " 180 " 190 " 1 " 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20... ... ... ... ... ... ... ... ... Similarly each century (200s, 300s, etc) contains another 20 zeros in the same pattern. Since 9 + 20 + 20 + 20 + 20 + 11 = 100, the 100th zero is the 11th zero in the 500s. Referring to the diagram above, this will be in the number 509, hence (509). c Australian Mathematics Trust www.amt.edu.au 40