Westpac Maths Competition (2 Viewers)

[kurt.physics]

Yea.. I did. I just did odd numbers which ended in 9, starting from 1999. Then divided them by 3.. to find the answer. Didn't take long, it was like the 4th number down, or something like that.

Was there another way to work it out? ie Algebra?

With Q6.. I just used basic algebra.

I'm pretty sure there was, but i didn't do it

Its an odd number, so its in the form 2n + 1

It leaves a remainder of 2 when divided by 3

so 3(2n + 1) + 2 = m

It leaves a remainder of 4 when divided by 5

so 5(2n + 1) + 4 = q

or something like that and then you get a diophantine equation (i think). You solve that or some thing....


I am a bit rough on Diophantine equations and all that stuff because i got this book "Problem-Solving Via the AMC" on Wednesday afternoon (the day before the test) and didnt read much that afternoon (tired). The next morning i took it to school and for the first 30 minutes in the morning before the DEAR (start of class) bell rang i was just sifting through it. So that helped me with question 6, but i haven done much Diophantine equations so i could solve it.

I was to lazy to do trial and error or any other method! Although i got Q19 right as soon as i saw that so, yeh (Q19 is circle with square thing inside)

 

[robot rabbit]

Question 18 was down by trial and error as well, a bit like rol. I answered it but question 19 made me hit my head.

 

[kurt.physics]

Question 18 was down by trial and error as well, a bit like rol. I answered it but question 19 made me hit my head.

What about Q26 - the three right-angled triangles? I didnt have a clue. Did you?

 

[Iruka]

I'm pretty sure there was, but i didn't do it

Its an odd number, so its in the form 2n + 1

It leaves a remainder of 2 when divided by 3

so 3(2n + 1) + 2 = m

It leaves a remainder of 4 when divided by 5

so 5(2n + 1) + 4 = q

or something like that and then you get a diophantine equation (i think). You solve that or some thing....


I am a bit rough on Diophantine equations and all that stuff because i got this book "Problem-Solving Via the AMC" on Wednesday afternoon (the day before the test) and didnt read much that afternoon (tired). The next morning i took it to school and for the first 30 minutes in the morning before the DEAR (start of class) bell rang i was just sifting through it. So that helped me with question 6, but i haven done much Diophantine equations so i could solve it.

I was to lazy to do trial and error or any other method! Although i got Q19 right as soon as i saw that so, yeh (Q19 is circle with square thing inside)

I think what you're thinking of is called the Chinese Remainder Theorem.

Many of the maths comp questions are based on elementary number theory (like CRT), counting methods and Euclidean geometry.

 

[kurt.physics]

I think what you're thinking of is called the Chinese Remainder Theorem.

Many of the maths comp questions are based on elementary number theory (like CRT), counting methods and Euclidean geometry.

Yeh, i think thats it. As i said, i really didnt study :S

But i agree with you, everybody would be much much better at solving these problems if they did the mathematical enrichment series (http://www.amt.canberra.edu.au/mcya.html).

I would like to do it but my school doesnt offer it. It pretty much teaches you all you need to know for the AMC (Diophantine equations, CRT, Counting techniques, Euclidean geometry etc)

 

[robot rabbit]

If Q26 was the one with the three right angled triangles with the perimeter and some of the sides given then yeah, i had a fairly good idea on how to work it out. However i noticed in the last 5 minutes i used the wrong numbers to calculate it so....

I hit my head at my silly mistake.

 

[Iruka]

But i agree with you, everybody would be much much better at solving these problems if they did the mathematical enrichment series (http://www.amt.canberra.edu.au/mcya.html).

I would like to do it but my school doesnt offer it. It pretty much teaches you all you need to know for the AMC (Diophantine equations, CRT, Counting techniques, Euclidean geometry etc)

That's a pity that your school doesn't offer it. Can you do it by yourself via distance ed of something like that?

 

[Aquawhite]

I really don't care what I get in the maths test anymore... I will be happy with participation this year because it was that hard!... Hopefully I can make a credit from it somehow.

 

[nerdsforever]

That was the worst competition ever. haha I got a Distinction last year and now I bet I'll get a Participation The first ten questions were relatively easy, but the rest was so freaking hard! but until I finished the test, I figured out how to do five or so questions. Why couldnt I have been able to solve them during the test? I blame nothing but my ignorance

 

[nerdsforever]

If Q26 was the one with the three right angled triangles with the perimeter and some of the sides given then yeah, i had a fairly good idea on how to work it out. However i noticed in the last 5 minutes i used the wrong numbers to calculate it so....

I hit my head at my silly mistake.

Yeh, I could have gotten it right if I didnt forget to half the area of the triangle in the end. I'm so stupid

 

[nerdsforever]

Just looking over this, yes, its quite long, but in a sense its not. It took me 1 minute in the test to think this up and calculate, so its probably easier than trial and error.

For example, this method of not guessing would have been particularly useful for Q18) A number is less than 2008. It is odd, it leaves a remainder of 2 when divided by 3 and a remainder of 4 when divided by 5. What is the sum of the digits of the largest such number?

EDIT: Did anyone manage to answer this?

I got this question wrong, but....

I put 1979. 9 + 1 + 7 + 9 = 26

But I just found out that the answer is actually 1994!!!!!!!!! damn it.

Id be pretty lucky to get a credit this time I reckon

 

[rolror88]

I got this question wrong, but....

I put 1979. 9 + 1 + 7 + 9 = 26

But I just found out that the answer is actually 1994!!!!!!!!! damn it.

Id be pretty lucky to get a credit this time I reckon

Ummm 1994 isn't odd.. At least I don't think so lol.. That was the first answer I first got and I nearly put it as my answer.

 

[nerdsforever]

Ummm 1994 isn't odd.. At least I don't think so lol.. That was the first answer I first got and I nearly put it as my answer.

ohh thats right. I forgot about that. phewww.

 

[candytrip]

Oh my god I hate you all, looking through your working I understand nothing. NOTHING!!!

I looked at the first question, the subtraction one, and thought "...fuck." After Q10, like just forget it. Hello, participation.

Someone worked out the first of the last 5 questions - the 3 triangles/area/perimetre question - it was 582. I hate simultaneous equations.

 

[kurt.physics]

the 3 triangles/area/perimetre question - it was 582. I hate simultaneous equations.

I got 582 as well... if i only got it in the test! i am hitting my head against the wall, just then it took me 3 minutes to fully work it out!

Also, for Question 28

The number 2008! (factorial 2008) means the product of all the integers 1, 2, 3, 4, ... , 2007, 2008. With how many zeroes does 2008! end?

I got 401 zeroes, did anyone get anything this or something else? It took me about 10-15 minutes to figure it out, basically, through finding the factorial of all the numbers from 0 to 16, gathered that there is the pattern of 0 zeroes at the end of 0-4, then 1 zero at the end of 5-9, and 2 zeroes at the end of 10-14 etc so you just have to divide 2008 by 5 to get 401.6, so there are 401 zeroes at the end.

I don't even understand Q27!

 

[ctpengage]

I got 582 as well... if i only got it in the test! i am hitting my head against the wall, just then it took me 3 minutes to fully work it out!

Also, for Question 28

The number 2008! (factorial 2008) means the product of all the integers 1, 2, 3, 4, ... , 2007, 2008. With how many zeroes does 2008! end?

I got 401 zeroes, did anyone get anything this or something else? It took me about 10-15 minutes to figure it out, basically, through finding the factorial of all the numbers from 0 to 16, gathered that there is the pattern of 0 zeroes at the end of 0-4, then 1 zero at the end of 5-9, and 2 zeroes at the end of 10-14 etc so you just have to divide 2008 by 5 to get 401.6, so there are 401 zeroes at the end.

I don't even understand Q27!

m8 big problem with ur logic. u realise that numbers such as 25 have a double factor of 5, then numbers such as 125 have 3 factors of 5, and so and so. The actual answer is 500. when the answer is nice whole and clean in these questions u pretty much no u got em rite

 

[hereiam]

that test basically killed my brain, i had to stare at empty space to recoever after that...
for the last ten questions i basically put C for all of them thinking "there has to be at least one thats a C"

 

[bored of sc]

that test basically killed my brain, i had to stare at empty space to recoever after that...
for the last ten questions i basically put C for all of them thinking "there has to be at least one thats a C"

Haha, I did the same thing but did (B) and the number 328 for the last 5. Hopefully we can each get one by chance!

 

[robot rabbit]

Just wondering can anyone beat this potential silly mistake.

I forgot that each angle of an equaliteral triangle is 60 degrees.

 

[x jiim]

If the farmer one was the one with the rain tanks, the answer was 10kL from the small tank to the big one. Some weird thing with ratios that I nearly screwed up ==
Think I did ok. If everyone else screwed up, that is. It's all relative...