prep数学考题求解

HDcoffee

[attachimg]89553[/attachimg]

VivienneLynn

还有一道题，为什么已经知道了mean是7，median是9，而且只有三个盒子，那么median怎么会是9呢

-- by 会员 HDcoffee (2011/8/18 10:44:00)

this one should be 3,



since the median is 9, mean is 7.

we could get the total of the 3 boxes is 7*3=21


the order of the 3 boxes could be A , 9, B


the weight of B should be 9 and more,
thus the maximum of A is 21-9-9=3.


3 should be the answer.

VivienneLynn

don't know the 3rd Q. I also chose C when I did the prep test, but I was guessing. Sb help.

sdcar2010

For #3,

If n is not divisible by 2, then both n-1 and n+1 are even numbers. Moreover, one of n-1 and n+1 has a factor of 4. So (n+1)*(n-1) has a factor of 2*4=8.

If n is not divisible by 3, then one of n-1 and n+1 has to be divisible by 3 (n-1, n, n+1 are consecutive numbers). Therefore (n-1)(n+1) has a factor of 3.

CCCCCCCCCCCCCCCCCCCCc

VivienneLynn

For #3,

If n is not divisible by 2, then both n-1 and n+1 are even numbers. Moreover, one of n-1 and n+1 has a factor of 4. So (n+1)*(n-1) has a factor of 2*4=8.

If n is not divisible by 3, then one of n-1 and n+1 has to be divisible by 3 (n-1, n, n+1 are consecutive numbers). Therefore (n-1)(n+1) has a factor of 3.

CCCCCCCCCCCCCCCCCCCCc

-- by 会员 sdcar2010 (2011/8/18 11:29:45)

thanks.


BTW, thanks for all your logic posts, I spent 3 days on them, and they really help a lot. THANKS.

HDcoffee

For #3,

If n is not divisible by 2, then both n-1 and n+1 are even numbers. Moreover, one of n-1 and n+1 has a factor of 4. So (n+1)*(n-1) has a factor of 2*4=8.

If n is not divisible by 3, then one of n-1 and n+1 has to be divisible by 3 (n-1, n, n+1 are consecutive numbers). Therefore (n-1)(n+1) has a factor of 3.

CCCCCCCCCCCCCCCCCCCCc

-- by 会员 sdcar2010 (2011/8/18 11:29:45)

HDcoffee

For #3,

If n is not divisible by 2, then both n-1 and n+1 are even numbers. Moreover, one of n-1 and n+1 has a factor of 4. So (n+1)*(n-1) has a factor of 2*4=8.

If n is not divisible by 3, then one of n-1 and n+1 has to be divisible by 3 (n-1, n, n+1 are consecutive numbers). Therefore (n-1)(n+1) has a factor of 3.

CCCCCCCCCCCCCCCCCCCCc

-- by 会员 sdcar2010 (2011/8/18 11:29:45)

thanks.


BTW, thanks for all your logic posts, I spent 3 days on them, and they really help a lot. THANKS.

-- by 会员 VivienneLynn (2011/8/18 11:41:28)

HDcoffee

So r is Zero, right?

For #3,

If n is not divisible by 2, then both n-1 and n+1 are even numbers. Moreover, one of n-1 and n+1 has a factor of 4. So (n+1)*(n-1) has a factor of 2*4=8.

If n is not divisible by 3, then one of n-1 and n+1 has to be divisible by 3 (n-1, n, n+1 are consecutive numbers). Therefore (n-1)(n+1) has a factor of 3.

CCCCCCCCCCCCCCCCCCCCc

-- by 会员 sdcar2010 (2011/8/18 11:29:45)

VivienneLynn

Oh, so r is Zero!!!

-- by 会员 HDcoffee (2011/8/18 11:45:52)

yes~