# Top PDF 2017IMAS Second Round Middle Primary Division Full Solutions

### 2017IMAS Second Round Middle Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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### 2017IMAS Second Round Middle Primary Division Problems

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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### 2018IMAS Second Round Middle Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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### 2014IMAS Second Round Middle Primary Division Full Solutions

By observation, we discover that each of the number in the first set when divided by 3 will give a remainder of 1, each number in the second set when divided by 3 will give a remainder of 2 and each number in the third set when divided by 3 or when each of them is divided by 3 will give a remainder of 0. Hence, according to the meaning of the problem, we must form a three-digit number where the digit in each place is selected from each set and the sum of all the digits formed when divided by 3 will give a remainder of 0, that is; the arrangement of all the three-digit numbers that formulate is always divisible by 3. Hence, we just need to find all those three-digit numbers from the above that are divisible by 2; or the digit in the units place is an even number.

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### 2015IMAS Second Round Middle Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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### 2016IMAS Second Round Middle Primary Division Full Solutions

Since 2017 1008 1009 1009 1010 = + < + , suppose we take out 1009~2017, which is a total of 1009 numbers, we cannot ensure that there are three balls that will satisfy the given conditions.(10 points) So suppose we take 1010 balls out. Suppose the largest number out of the 1010 balls is M , then the difference between M and the number of other balls taken out has 1009 different values, and both are less than 2017.(5 points) Since there are only 1007 balls that have not been taken out, at least one difference M − x is the number y of the balls taken out, where x is the number of the removed ball. So x, y, x + = y M are taken out of the ball number which satisfies the condition above. (5 points)

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### 2017IMAS First Round Middle Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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### 2017IMAS Second Round Upper Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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### 2018IMAS First Round Middle Primary Division Full Solutions

【Solution】 It is obvious the model is contained in a cube of edge length 3 and contains some of the 27 unit cubes of the large cube. The unit cube in the center has to appear in the model. The cubes at the center of faces of the large cube must not appear. The cubes at the middle of edges of the large cube must not appear. Moreover, one of the two unit cubes at any adjacent corners of the large cube must appear. It is then obvious by pigeonhole principle, one needs at least 5 unit cubes. One such construction is in the figure below.

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### 2014IMAS First Round Middle Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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### 2015IMAS First Round Middle Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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### 2016IMAS First Round Middle Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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### 2018IMAS Second Round Upper Primary Division Full Solutions

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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### 2014IMAS Second Round Upper Primary Division Full Solutions

Answer: 45 number sentences 10. An ant starts from the top left corner square of a 3 5  chessboard. It moves from a square to the adjacent square in the same row or column. After visiting every square exactly once, it ends up at the square in the middle row second column from the right. How many different paths can the ant follow?

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### 2015IMAS Second Round Upper Primary Division Full Solutions

Answer: 3600 dollars. 7. The diagram shows a path consisting of four semi-circular arcs. Each arc is of length 100 m and uses a different side of a square as its diameter. Initially, Jane is at A and Yves at B. They start walking counter-clockwise at the same time. Jane’s speed is 120 m per minute and Yves’s is 150 m per minute. Each pauses for 1 second whenever they are at the points A, B, C or D. How many seconds after

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### 2016IMAS Second Round Upper Primary Division Full Solutions

shown, where the middle part is made up of a hollow rolled cardboard. The tissue paper is divided into several sheets and rolled over the cardboard. The pack indicates: “138 mm 100 mm × per sheet, 3 ply”, which means that each sheet is 138 mm long and 100 mm wide and it contains 3 layers. Given that each roll of tissue paper is 0.13 mm thick, the diameter of cardboard roll is 5 cm, and the diameter of the rolled tissue paper is 12 cm, how many sheets of tissue paper, rounding to integer, are there in a pack? (Take π = 3.14 )

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### 2017IMAS First Round Upper Primary Division Full Solutions

【Solution】 If there is no carry when the number is added by 1, sum of digits increases by 1, contradiction to that both sum of digits are divided by 4. If there is a carry, the last digit is 9. Since the sum of all digits changes by 8 when one carry happens and by 17 when two carries happens. It follows that only one carry should happen and the original number has the sum of all digits divided by 4. In order to take the number to be maximal, the first digit should also be 9. The second digit can be either 2 or 6. The largest one is 969.

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### 2018IMAS Second Round Middle Primary Division Problems

After making your choice, fill in the appropriate letter in the space provided. Each correct answer is worth 4 marks. There is no penalty for an incorrect answer. Questions 6 to 13 are a short answer test. Only Arabic numerals are accepted; using other written text will not be honored or credited. Some questions have more than one answer, as such all answers are required to be written down in the space provided to obtain full marks. Each correct answer is worth 5 marks. There is no penalty for incorrect answers.

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### 2014IMAS Second Round Middle Primary Division Problems

After making your choice, fill in the appropriate letter in the space provided. Each correct answer is worth 4 marks. There is no penalty for an incorrect answer. Questions 6 to 13 are a short answer test. Only Arabic numerals are accepted; using other written text will not be honored or credited. Some questions have more than one answer, as such all answers are required to be written down in the space provided to obtain full marks. Each correct answer is worth 5 marks. There is no penalty for incorrect answers.

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### 2015IMAS Second Round Middle Primary Division Problems

Notice: Individual students, nonprofit libraries, or schools are permitted to make fair use of the papers and its solutions. Republication, systematic copying, or multiple reproduction of any part of this material is permitted only under license from the Chiuchang Mathematics Foundation.

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