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**tortoise****Guest**

Can you help me solve these problems?

www.nus.edu.sg/oam/doc/uee/math.pdf

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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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**tortoise****Guest**

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.

**liuv****Member**- Registered: 2006-05-14
- Posts: 29

DO you have msn? i can solve some problems.

I'm from Beijing China.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,490

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....:)

Character is who you are when no one is looking.

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**tortoise****Member**- Registered: 2006-05-20
- Posts: 1

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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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,535

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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