All About Math Olympiad Training & Questions

[r2010]

verykiasu2010:

mjl:

I read the below from the NMOS 2011 webpage:

\"NUS High School is officially appointed by the Gifted Education Branch to be the Lead Agency for the 8th International Mathematics and Science Olympiad for Primary Schools (IMSO). For 2011, IMSO will be held on 2nd Sep to 6th Sep in the Philippines.

The top 20 individual winners of NMOS will be shortlisted to join the training squad for IMSO 2011. Within the training squad, a final team of 6 official participants and 2 reserves will be selected to form the Mathematics team for IMSO 2011. Only the 6 official participants are eligible to travel to the Philippines. \"

I think in the past, the participants from IMSO come from the winners of RI/RGS Math and Science competitions for P5 students. With the change, I wonder if RI/RGS will still be organising Math and Science competitions for P5.

the fight for top students to join the sec school has just intensified

You mean the school will try to get all these students in their school by DSA?

[r2010]

cchew:

tisha:

Where can we get the past questions of RI olympiad? TIA

Preparatory Course For Math Contest

In order to help students prepare for the contest in 2011, Raffles Institution and HeyMath! have developed a set of preparatory lessons and worksheets to enable students expand their knowledge and improve problem solving skills necessary for success. The preparatory course focuses on mathematical reasoning and includes useful approaches and methods to tackle problems.

This course also includes the question papers used in rounds 1 and 2 of the Raffles Institution Primary Mathematics World Contest (RIPMWC) in 2008, 2009 & 2010.

http://www.heymath.com/web/products/productMathContest.jsp

I went to the website.. Has anyone buy the programme? Is it easy to understand it online?

[cchew]

r2010:

Do the school publish who are the representive for IMSO for past years.. just wondering which school is producing all these Maths wizard.. :?

Refer for here for last year IMSO
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=11737&start=10&postdays=0&postorder=asc&highlight=

[zsqchx]

hi, could you help me to solve this question. Thanks


There is a big migic fruit tree in a garden. It bears fruits at a constant rate every day. Man will pick the fruits and sell. Animals will pluck and eat the fruits. 25 men can pick all the fruits in 15 days and 66 animals can eat all the fruits in 10 days. If 1 man can pick twice the number of fruits as 1 animal, find the number of days it take for 10 men to pick and 10 animals to eat all the fruits together?

[FrekiWang]

zsqchx:

hi, could you help me to solve this question. Thanks


There is a big migic fruit tree in a garden. It bears fruits at a constant rate every day. Man will pick the fruits and sell. Animals will pluck and eat the fruits. 25 men can pick all the fruits in 15 days and 66 animals can eat all the fruits in 10 days. If 1 man can pick twice the number of fruits as 1 animal, find the number of days it take for 10 men to pick and 10 animals to eat all the fruits together?

the question is equivalent to:
constant rate of growth
25men can pick in 15days
33men can pick in 10days
and you are asked to find how many days will it take if there are 15men picking.

25x15=375 men day, which is equivalent to the original amount + the growth in 15 days.

33x10=330 menday, which is equivalent to the original amount+ the growth in 10 days.

375-330=45 menday, which is then equivalent to the growth in 5 days.

45/5=9 men, which means the rate of growth is equivalent to the amount picked by 9 men per day.

The original amount on the tree is euivalent to
375 - 9 x 15 = 240 (or 330 - 9 x 10 =240) menday

now there are 15 men picking, the growth rate is 9 men, that means 15 - 9=6 men is deducted from the original amount per day.

So, it will take 240/6=40days - answer

[zsqchx]

Much appreciate your time and help!

[Steven_Phua]

Hi， please kindly help me for this:


Given that a and b are both positive integers, prove that (a-b)^2 is divisible by (4ab-1) only if a=b

(a-b)^2 here means the square of (a-b)

[FrekiWang]

Hi， please kindly help me for this:


Given that a and b are both positive integers, prove that (a-b)^2 is divisible by (4ab-1) only if a=b

(a-b)^2 here means the square of (a-b)

Hi, sorry for the late reply, was quite busy in the afternoon.

Here is my solution:

Proof:

Suppose there are some solutions where a<>b, since the expressions are symmetrical of a and b, we can assume a>b.

Let (a0,b0) be the solution pair which makes a0+b0 the least.

Let (a-b)^2/(4ab-1)=k for this pair, where k is some positive interger, move the denominator to the RHS, expand the brackets and make every in the form of a quadratic equation in a, we have,
a^2 - (2b+4bk)a + (b^2+k) = 0, there are two roots, one of them is a0, the other root a1 is (2b0 + 4b0k -a0) (sum of roots). Obviously a1 is also a positive integer as their product b^2+k is positive

Since (a0,b0) is the solution pair which minimizes a+b, a0+b0<=a1+b0
i.e. a0<=a1
The product of roots = b0^2+k=a0a1>=a0^2.
therefore k>=a0^2-b0^2
i.e. (a0-b0)^2/(4a0b0-1)=k>=(a0+b0)(a0-b0)
Multiplying both sides by the positive (4a0b0-1), we have
(a0-b0)^2>=(a0+b0)(a0-b0)(4a0b0-1)
since a0>b0, dividing both sides by the positive (a0-b0), we have
(a0-b0)>(a0+b0)(4a0b0-1)

Since a0>b0>=1,
thus
0<a0-b0<a0+b0
0<1<4a0b0-1
the product (a0-b0)*1<(a0+b0)(4a0b0-1) which leads to a contradiction to the conclusion above. Hence, there is no solution pair for a>b, similar neither there is for a<b, so the only possible pairs occur when a=b

[SKT]

Let S = {1,2,…,4022}. Let T be the subset of S such that no number in T is divisible by another. What is the maximum size of T, i.e. the maximum number of elements that can be in T?


TIA

[FrekiWang]

SKT:

Let S = {1,2,...,4022}. Let T be the subset of S such that no number in T is divisible by another. What is the maximum size of T, i.e. the maximum number of elements that can be in T?

TIA

The answer is 2011, i.e. half of 4022.

A specific example can be T={2012,2013,2014,....,4022} which obviously satisfies the condition.

Now let's prove the maximum size cannot exceed 2011.

Define 2011 subsets of T as follows:

A1={1, 1x2, 1x2x2, 1x2x2x2 ,...., 1x2^11}
(1x2^11=2048, 1x2^12=4096>4022)

A3={3, 3x2, 3x2x2, ...., 3x2^10}

A5={5, 5x2, 5x2x2, ...., 5x2^9}

A7={7, 7x2, 7x2x2, ...., 7x2^9}
....
A2009={2009, 2009x2}

A2011={2011, 2011x2}

A2013={2013}

A2015={2015}
....
A4019={4019}

A4021={4021}

A little explaination here for those who cannot see the pattern,
Subset A(2k-1) (k=1,2...2012) contains all numbers that are equal to (2k-1) multiply by powers of 2 and do not exceed 4022. There are 2011 such subsets in total.

Suppose the maximum size of the subset T exceeds 2011 (i.e. at least 2012) and all elements belong to at least one of the 2011 subsets mentioned above, then according to 'pigeonhole principle', at least 2 or more elements must belong to one of the subsets.

For these 2 elements, the greater number is always divisible by the smaller one, thus it is not possible for a subset S to have a size that exceeds 2011.

PS: the problem can be extended to:

Let S = {1,2,...,2n}. Let T be the subset of S such that no number in T is divisible by another. What is the maximum size of T, i.e. the maximum number of elements that can be in T?

and the answer is n, you can show n+1 is not possible using a similar approach, in fact this is one classical application of 'pigeonhole principle'