Top PDF 2014IMAS First Round Upper Primary Division Full Solutions

2014IMAS First Round Upper Primary Division Full Solutions

2014IMAS First Round Upper Primary Division Full Solutions

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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2014IMAS First Round Upper Primary Division Problems

2014IMAS First Round Upper 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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2017IMAS First Round Upper Primary Division Full Solutions

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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2016IMAS First Round Upper Primary Division Full Solutions

2016IMAS First 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 Upper Primary Division Full Solutions

2018IMAS First Round Upper Primary Division Full Solutions

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.

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2015IMAS First Round Upper Primary Division Full Solutions

2015IMAS First 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

2014IMAS 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 First Round Middle Primary Division Full Solutions

2014IMAS First Round Middle Primary Division Full Solutions

【Note】By symmetry property of a diagram, We just need to count the number of sides of the upper half and then be multiplied by 2. 6. Thirty students numbered from 1 to 30 stand in a row. The teacher announces, “Will those numbered from 1 to 10 inclusive take one step forward, and those numbers 20 to 30 inclusive take one step backward.” How many students remain in place?

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2018IMAS Second Round Upper Primary Division Full Solutions

2018IMAS Second Round Upper Primary Division Full Solutions

(5 points) Construct 90 codes as follows: the first two digits take 10 to 99, once for each; the third digit equals the last digit of sum of first two digits. (5 points) Next, we show that such set of codes satisfies the command. For any code abd , a + − b d is divisible by 10, thus each of a , b , d is determined by the other two. If two codes have identical digits at two corresponding places, their digits at the third place are also the same, showing that they are the same code. Thus every two codes have identical digits at no more than one places. (10 points)
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2017IMAS Second Round Upper Primary Division Full Solutions

2017IMAS Second Round Upper Primary Division Full Solutions

If there are less than 4 rows with pieces more than 1, covering 4 rows with the most pieces will leave all remaining row each contains no more than one piece. The remaining pieces can be covered by 4 columns, contradiction. (5 marks) Put 13 pieces in the squares as the picture below will ensure that any 4 rows and 4 columns can not cover all pieces. (5 marks) Noticing that pieces in the upper left

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2015IMAS Second  Round Upper Primary Division Full Solutions

2015IMAS Second Round Upper Primary Division Full Solutions

【Suggested Solution】 First take away one bottle of each kind, so that the relevant total capacity is now reduced to 16. If we use an equal number of 0.4 L bottles and 0.6 L bottles, the number of 0.6 L bottles can be any number from 0 to 16 inclusive. Since three 0.4 L bottles can be replaced by two 0.6 L bottles and we have up to sixteen 0.4 L bottles, enough for five replacements, the number of 0.6 L bottles can stretch to16+5×2=26.

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2016IMAS Second Round Upper Primary Division Full Solutions

2016IMAS Second Round Upper Primary Division Full Solutions

10. Alex and Charles both sent parcels of weight exceeding 10 kg . The postage rates Alex and Charles were both sending parcels. The postage rates are as follows: For the first 10 kg and below, the postage price is $6 per kg; the postage price for each successive kilogram after 10 kg is $2 lower than the original price. Given that Alex’s parcel is 20% heavier than Charles’ and his postage price is $12 more than that of Charles, find the weight, in kg, of the parcel sent by Alex?

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2018IMAS First Round Middle Primary Division Full Solutions

2018IMAS 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 First Round Middle Primary Division Full Solutions

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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2015IMAS First Round Middle Primary Division Full Solutions

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

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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2014IMAS Second  Round Middle Primary Division Full Solutions

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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2016IMAS First Round Upper Primary Division Problems

2016IMAS First Round Upper 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 First Round Upper Primary Division Problems

2018IMAS First Round Upper 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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2017IMAS First Round Upper Primary Division Problems

2017IMAS First Round Upper 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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