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「Answer」AMC8 -- 2012年真题解析与视频讲解( 二 )


「Answer」AMC8 -- 2012年真题解析与视频讲解
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Answer: D
Solution:
Notice that the mean of this set of numbers, in terms of x, is: (3+4+5+6+6+7+x)/7=(31+x)/7.Because x should be a whole number, thus x could be 4, 4+7, 4+7*2…. Because we know that the mode must be 6 (it can't be any of the numbers already listed, as shown above, and no matter what x is, either 6 or a new number, it will not affect 6 being the mode), and we know that the mode must equal the mean, we can set the expression for the mean and 6 equal: (31+x)/7=6 , =>x=11.Problem 12
「Answer」AMC8 -- 2012年真题解析与视频讲解
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Answer: A
Solution:
3^1=>3; 3^2=>9; 3^3=>7; 3^4=>1; 3^5=>3…We observe that there is a pattern for the units digit which recurs every four powers of three. Using this pattern, the remainder of 2012/4 is 0, thus, the units digit of 3^2012 is 1. The units digit of 13^2012 is 1 too.Problem 13「Answer」AMC8 -- 2012年真题解析与视频讲解
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Answer: C
Solution:
We assume that the price of the pencils remains constant. Since the price of the pencils is more than one penny, we can find the price of one pencil (in cents) by taking the greatest common divisor of 143 and 187, which is 11. Therefore, Jamar bought 143/11=13 pencils, d Sharona bought 187/11=17 pencils. Thus, Sharona bought 17-13=4 more pencils than Jamar.Problem 14
「Answer」AMC8 -- 2012年真题解析与视频讲解
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Answer: B
Solution:
This problem is very similar to a handshake problem. We use the formula n(n-1)/2 to usually find the number of games played (or handshakes). Now we have to use the formula in reverse.So we have the equation n(n-1)/2=21 . Solving, we find that the number of teams in the BIG N conference is 7.Problem 15「Answer」AMC8 -- 2012年真题解析与视频讲解
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Answer: D
Solution:
To find the answer to this problem, we need to find the least common multiple of 3,4,5,6. The least common multiple of the four numbers is 60, and by adding 2, we find that that such number is 62.Problem 16
「Answer」AMC8 -- 2012年真题解析与视频讲解
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Answer: C
Solution:
In order to determine the largest number possible, we have to evenly distribute the digits when adding. The two numbers that show an example of this are 97531 and 86420. The digits can be interchangeable between numbers because we only care about the actual digits.Only C is OK .Problem 17「Answer」AMC8 -- 2012年真题解析与视频讲解
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Answer: B
Solution:
The first answer choice A(3), can be eliminated since there must be 10 squares with integer side lengths. We then test the next smallest side length which is 4. The square with area 16 can be partitioned into 8 squares with area 1 and two squares with area 4, which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is 4.Problem 18
「Answer」AMC8 -- 2012年真题解析与视频讲解
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Answer: A
Solution:
The problem states that the answer cannot be a perfect square or have prime factors less than 50. Therefore, the answer will be the product of at least two different primes greater than 50. The two smallest primes greater than 50 are 53 and 59. Multiplying these two primes, we obtain the number 3127, which is also the smallest number on the list of answer choices.Problem 19「Answer」AMC8 -- 2012年真题解析与视频讲解
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Answer: C
Solution:
Let n be the number of marbles, there are n-6 red marbles, n-8 green marbles,n-4 blue marbles. At the same time (n-6)+(n-8)+(n-4)=n. thus, n=9.Problem 20


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