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#1 2006-05-19 16:33:57
- tortoise
- Guest
Help me! Thanks!
Can you help me solve these problems?
www.nus.edu.sg/oam/doc/uee/math.pdf
#2 2006-05-19 16:39:36
- Ricky
- Moderator
- Registered: 2005-12-04
- Posts: 3,791
Re: Help me! Thanks!
Do you have any specific ones that you are having trouble on? Have you tried to do them yet?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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#3 2006-05-19 16:48:43
- tortoise
- Guest
Re: Help me! Thanks!
Yeah, thanks for quick reply, Mod!
I'm a Vietnamese and I intend to apply for an exam call Entrance Examination, of National University of Singapore. As you see, the problems are written in English and too difficult for me to do (except some easy problem which I can solve).
I hope that someone would help me to solve these problems.
Thanks in advance.
#4 2006-05-19 17:00:58
- liuv
- Member
- Registered: 2006-05-14
- Posts: 29
Re: Help me! Thanks!
DO you have msn? i can solve some problems.
I'm from Beijing China.
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#5 2006-05-19 17:04:16
- Jai Ganesh
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- Registered: 2005-06-28
- Posts: 48,388
Re: Help me! Thanks!
Some answers....
2. (B) Given f(x)=x²+1 is x is greater than or equal to 2. In this case, x=2.
4. (D) Use the formula for n(n+1)/2 for n=500 and then subtract 71/2{14+490}from it.
19.(C)
I shall try the others later....:)
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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#6 2006-05-20 19:35:00
- tortoise
- Member
- Registered: 2006-05-20
- Posts: 1
Re: Help me! Thanks!
Thank all of you!
Problems 2 is easy and I can solve them. But other problems left, I can't......
Problem 4: how do you form that? Pls.
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#7 2006-05-20 23:07:30
- MathsIsFun
- Administrator
- Registered: 2005-01-21
- Posts: 7,713
Re: Help me! Thanks!
Sorry, Ganesh, I couldn't resist attempting it:
4: The sum of all positive integers less than or equal to 500 which are not multiples of 7
Firstly, the sum of 1 to 500 is (1+500)/2 (the average) times 500 = 125250
Then, all multiples of 7 are 7 to 497, and there are 71 of them. Their sum is (497+7)/2 (the average) times 71 = 17892
125250 - 17892 = 107358
I think 
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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