# Top PDF 2017IMAS First Round Junior Division Problems ### 2017IMAS First Round Junior Division Problems

When your teacher gives the signal, begin working on the problems. I I nt n te er rn na at t io i on n al a l M Ma at t he h em ma at ti i cs c s A A ss s se es ss sm me en n ts t s f fo or r S Sc ch ho oo ol ls s 2017 JUNIOR DIVISION FIRST ROUND PAPER Time allowed：75 minutes

9 閱讀更多 ### 2016IMAS First Round Junior Division Problems

9 閱讀更多 ### 2018IMAS First Round Junior Division Problems

When your teacher gives the signal, begin working on the problems. I I nt n te er rn na at t io i on n al a l M Ma at t he h em ma at ti ic cs s A A ss s se es ss sm me en n ts t s f fo or r S Sc ch ho oo ol ls s 2018 JUNIOR DIVISION FIRST ROUND PAPER Time allowed：75 minutes

9 閱讀更多 ### 2014IMAS First Round Junior Division Problems

9 閱讀更多 ### 2015IMAS First Round Junior Division Problems

（A）20 （B）31 （C）42 （D）53 （E）64 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?

9 閱讀更多 ### 2017IMAS Second Round Junior Division Problems

z Do not open the contest booklet until you are told to do so. z Be sure that your name and code are written on the space provided above. z Round 2 of IMAS is composed of three parts; the total score is 100 marks. z Questions 1 to 5 are given as a multiple-choice test. Each question has five possible options marked as A, B, C, D and E. Only one of these options is correct.

9 閱讀更多 ### 2017IMAS First Round Junior Division Full Solutions

（A）12 （B）18 （C）24 （D）30 （E）36 【Solution 1】 From the given information, one student will participate in two events and each of other two students participate on exactly one event. Let the student (the one will participate in 2 events) choose the events that he wants to register, then there are 4 ways of registering the two events, follow by the first student with one event has 3 ways to register his event, the second student with one event has also 3 ways to choose his event. Thus, the total number is 3 4 3 × × = 36 ways of participating.

10 閱讀更多 ### 2017IMAS First Round Middle Primary Division Problems

9 閱讀更多 ### 2017IMAS First Round Upper Primary Division Problems

14. There are two routes starting in a bus stop. A bus departs for the first route every 8 minutes and departs the second route every 10 minutes. At 6:00 in the morning, two buses depart for the two routes at the same time. Among the choices below, when will the buses depart for the two routes simultaneously?

9 閱讀更多 ### 2018IMAS Second Round Junior Division Problems

Do not open the contest booklet until you are told to do so. Be sure that your name and code are written on the space provided above. Round 2 of IMAS is composed of three parts; the total score is 100 marks. Questions 1 to 5 are given as a multiple-choice test. Each question has five possible options marked as A, B, C, D and E. Only one of these options is correct.

9 閱讀更多 ### 2014IMAS Second Round Junior Division Problems

Do not open the contest booklet until you are told to do so. Be sure that your name and code are written on the space provided above. Round 2 of IMAS is composed of three parts; the total score is 100 marks. Questions 1 to 5 are given as a multiple-choice test. Each question has five possible options marked as A, B, C, D and E. Only one of these options is correct.

9 閱讀更多 ### 2015IMAS Second Round Junior Division Problems

Do not open the contest booklet until you are told to do so. Be sure that your name and code are written on the space provided above. Round 2 of IMAS is composed of three parts; the total score is 100 marks. Questions 1 to 5 are given as a multiple-choice test. Each question has five possible options marked as A, B, C, D and E. Only one of these options is correct.

7 閱讀更多 ### 2016IMAS Second Round Junior Division Problems

z Do not open the contest booklet until you are told to do so. z Be sure that your name and code are written on the space provided above. z Round 2 of IMAS is composed of three parts; the total score is 100 marks. z Questions 1 to 5 are given as a multiple-choice test. Each question has five possible options marked as A, B, C, D and E. Only one of these options is correct.

9 閱讀更多 ### 2016IMAS First Round Junior Division Full Solutions

3 2 5 2 2016 − 2016 = 2016 (2016 × − = 1) 2015 2016 2017 × × = × × × × × × 2 3 5 7 13 31 2017 , so the prime factors of the positive integer which satisfy the conditions should be 11, 17, 19, …, 29, 37, …, 2011, 2027, …. Since the positive integer has 12 positive divisors, it is of the form p 11 , p q 5 , p q 3 2 or p qr 2 , where p, q and r are different prime numbers. It is obviously that the first three forms are all greater than 10 5 and the smallest value of the last one is 11 2 × × = 17 19 39083 10 < 5 .

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If 2 and 8 are both to the right of 4, the sequence partially looks like 4ab2c8, where 1 must be a. If 4 is the first from the left, then the sequence looks like 41b2c8de, 3 and 6 must be at b, d, two choices for them and two choices for remaining 5 and 7;

11 閱讀更多 ### 2014IMAS First Round Junior Division Full Solutions

From the given information, it shows the sum of the three numbers in the second row is equal to twice the sum of the three numbers in the first row, it follows that the difference of the sum of three numbers in the second row and the sum of three numbers in the first row equal the sum of three numbers in the first row, that is; the difference of these two rows is 19 + 20－5－6 = 28, then we have ☆ = 28－5－6 = 17. Thus, we select option (D).

11 閱讀更多 ### 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?

8 閱讀更多 ### 2017IMAS Second Round Junior Division Full Solutions

【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 .

8 閱讀更多 ### 2016IMAS First Round Upper Primary Division Problems 