# 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 ...

9 ### 2016IMAS First Round Junior Division Problems

9 ### 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 ...

9 ### 2014IMAS First Round Junior Division Problems

9 ### 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 ...

9 ### 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 ...

9 ### 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 ...

10 ### 2017IMAS First Round Middle Primary Division Problems

9 ### 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 ...

9 ### 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 ...

9 ### 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 ...

9 ### 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 ...

7 ### 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 ...

9 ### 2016IMAS First Round Junior Division Full Solutions

... 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, ...the first three forms are ...

11 ### 2018IMAS First Round Junior Division Full Solutions

... 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 ...

11 ### 2014IMAS First Round Junior Division Full Solutions

... 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 ...

11 ### 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 ...

8 ### 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. ...

8 ### 2016IMAS First Round Upper Primary Division Problems 