Top PDF 2017IMAS First Round Middle Primary Division Full Solutions

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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.

【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.

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.

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.

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.

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.

【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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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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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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Totally there are 20  20 8 4 1 53     isosceles right triangles. Hence (E). Answer： （E） 21. Mike constructs a sequence in the following way: the first two terms are 1 and 2. Starting from the third term, each term is the smallest possible integer that is not relatively prime to the previous term and has not yet appeared in any of the previous terms. Find the 20 th term of this sequence.

Since 2014 = 335 × 6 + 4, it means that in the first 2014 terms in the sequence, there are 335 complete periods with an incomplete period of four remaining terms. Therefore, there are 335 terms in the sequence that are divisible by 4. Answer: 335 22. The inside lane of a track has length 400 m and the outside lane has length less than 500 m. From the marked line, as shown in the diagram, Max and Lynn start running counterclockwise along the track at the same time. Max runs at constant speed on the inside lane. Lynn, whose constant speed is 3 times that of Max, runs on the outside lane. The first time both are back together at the marked line, Max has completed 3 laps. What is the length, to the nearest m, of the outside lane?

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its last two digits will give a number divisible by 6. How many "lucky" numbers are there? 【 Solution 】 Being divisible by 6 is equivalent to being divisible by both 2 and 3. Thus, the last two digits of a lucky number is even. A number is divisible by 3 if and only if the sum of all digits is divisible by 3. The remainder divided by 3 of the first number of the lucky number is determined by the last two digits. For any remainder, there are exactly 3 non-zero digits. Thus, the number of all lucky numbers is 5 5 3 × × = 75 .

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.

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.

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.