# Top PDF 2017IMAS First Round Junior Division Full Solutions

### 2017IMAS First Round Junior 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 Junior 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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### 2016IMAS First Round Junior 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 Junior 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 Junior 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 Junior Division Full Solutions

Subtracting this from the total surface area of the individual cubes, we have 14 6 42    42 . Answer： （C） 19. In an election between four candidates, they are supported respectively by 11, 12, 13 and 14 of the first 50 voters. Six more votes are to be cast, each for one of the four candidates. In how many ways can the candidate currently with 13 supporters become the uncontested winner?

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### 2017IMAS Second Round Junior 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 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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### 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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### 2018IMAS Second Round Junior 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 Junior 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 Second Round Junior 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 Junior Division Full Solutions

Thus, ∠ BFC = ∠ AFE = 180 ° − ° − ° = ° 90 35 55 . Answer： （A） 3. 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; for each successive kilogram after 10 kg, the postage price per kg is slightly lower than that of the first 10kg. It is known that the weight of Alex’s parcel is 20% heavier than Charles’ parcel, and that the postage prices for Alex and Charles are \$92 and \$80 respectively. How much more is the postage price per kg of the first 10 kg than that of each succeeding kg above 10 kg?

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### 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 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 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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### 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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### 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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### 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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### 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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