刚做完prep1。结局惨烈。几个数学题。求战友们解答。

xtxtxtli · 发表于 2011-1-5 11:20:01

1. for all positive integers m,
m（包了一个方框）=3m， if m is odd;
m(包了一个方框）=1/2m, if m is even
which is equivalent to 带方框的9*带方框的7？

我算的是81，答案是27

2.if n is a positive integer and the product of all the integers from 1 to n, is a multiple of 990, what is the least possible value of n?

答案:11

3. of the 60 animals on a certain farm, 2/3 are either pigs or cows. how many of the animals are cows?
1) the farm has more than twice as many cows as it has pigs.
2) the farm has more than 12 pigs.

答案选c

4. an integer greater than 1 that is not prime is called composite. if the two-digit integer n is greater than 20, is n composite?
1) the tens digit of n is a factor of the units digit of n.
2) the tens digit of n is 2.

选A。

5. is z equal to the median of the three positive integers x,y, and z?
1) x<y+z
2) y=z

一模数学46分。错了13题。请教距离51分还有多远呢？
对各位朋友的帮助，不胜感激！！

sdcar2010 · 发表于 2011-1-5 11:39:47

1. for all positive integers m,
m（包了一个方框）=3m， if m is odd;
m(包了一个方框）=1/2m, if m is even
which is equivalent to 带方框的9*带方框的7？

First, I believe, the nubmer 7 should be number 6.

Secondly, I believe the answers are all in blocks! Think about the ramifications!

sdcar2010 · 发表于 2011-1-5 11:42:35

2.if n is a positive integer and the product of all the integers from 1 to n, is a multiple of 990, what is the least possible value of n?

n! = n*(n-1)*. . . *2*1= m*990 = m*11*9*5*2

So n! contains the factors of 11, 9, 5, and 2. Then n is at least as big as 11!

sdcar2010 · 发表于 2011-1-5 11:49:49

3. of the 60 animals on a certain farm, 2/3 are either pigs or cows. how many of the animals are cows?
1) the farm has more than twice as many cows as it has pigs.
2) the farm has more than 12 pigs.

So among the 60 animals are 40 pigs and cows.

1) number of cows > 2*[numbers of pigs]. Two many possibilities. 39 cows and 1 pig; 38 cows and 2 pigs; etc. Insufficient.
2) number of pigs > 12. Again two many possibilities. 39 pigs and 1 cows; 38 pigs and 2 cows, etc. Insufficient.

1) and 2). Based on 2) At least 13 pigs. Then according to 1), at least 27 cows. But wait. 13+ 27 = 40. That's it. That's the only possibility. Sufficient.

sdcar2010 · 发表于 2011-1-5 11:56:31

4. an integer greater than 1 that is not prime is called composite. if the two-digit integer n is greater than 20, is n composite?
1) the tens digit of n is a factor of the units digit of n.
2) the tens digit of n is 2.

1) Let's say the units digit of n is a. Then the tens digit of n is a factor of a. Therefore, n = 10*m*a + a = a*(10m +1). Then n has a factor of a and a is bigger than 1 since n is greater than 20. n is a composite for sure. Sufficient.
2) Insufficient. 21, 22, 23 ....

sdcar2010 · 发表于 2011-1-5 11:58:42

5. is z equal to the median of the three positive integers x,y, and z?
1) x<y+z
2) y=z

1) Insufficient. x=1, y=2, z=3. x=1, y=3, z=2.
2) Sufficient. Either x, y=z; or y=z; x.

sdcar2010 · 发表于 2011-1-5 22:39:41

If you can see the attached picture, you would have a better time to solve the problem 1 LZ posted.

According to the stimulus, [9] = 3*9 =27. [6] = (1/2) * 6 = 3. So [9]*[6] = 27 * 3 = 81. So far so good. But if you choose (a) as the correct answer, you were wrong because (a) is [81] = 3*81 = 243!

On the other hand, (d) [27] = 3*27 = 81, is the right answer!

Rebeca · 发表于 2011-1-5 22:45:03

N人！服了！！

gesteven · 发表于 2011-1-10 10:26:04

大侠回复太全了，佩服佩服。

Shine61 · 发表于 2011-3-3 17:45:13

2.if n is a positive integer and the product of all the integers from 1 to n, is a multiple of 990, what is the least possible value of n?

n! = n*(n-1)*. . . *2*1= m*990 = m*11*9*5*2

So n! contains the factors of 11, 9, 5, and 2. Then n is at least as big as 11!

-- by 会员 sdcar2010 (2011/1/5 11:42:35)

还是不太明白。

刚做完prep1。结局惨烈。几个数学题。求战友们解答。

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