What do I have to do to get into a really good college? Also, is what I am already doing enough?

Author: admin  //  Category: maths competitions

Right now I am a high school freshman, but I want to get into a REALLY good college when the time comes (like Harvard or something).

My grades for this year so far are 3 A+’s, two A’s, and a B+.

I play three sports (cross country, swimming, and track), though I do not excel particularly in either of them (however, our school’s swim team is very prestigious–do you think being on a very good swim team would look good for college, even if I myself am not a very fast swimmer?)

I go to a good school and am also fluent in a foreign language.

Would all of these things look good for college? Also, how good does simply playing a sport look for college–or would you have to be really extraordinary at a sport for it to actually count?

Should I try to participate more in math competitions and/or debate club? (Also, if I did, would I also have to excel at that for it to look good for college?)

Also, what other things should I do to better my chances of a good college?
and, in general, what are the requirements to get into really good colleges such as harvard, etc.?

any and all info would help! (in as much detail as possible please!)

thanx! =]
also, do your health and physical education grades count towards your GPA (not your "school GPA" but the one that colleges actually get to see?)

yah your fine, make sure to do good in your ACT tho

youll go thro that in your jr or senior yr

also concentrate your grades in your JR yr

Australia Maths Competition problem - need help!?

Author: admin  //  Category: maths competitions

Thirty-six 1×1x1 cubes are used to make a rectangular prism. How many different rectangular prisms can be made? If there’s a math formula in there, can you tell me what it is?

Considering only unique prisms and not rotations, there are eight.

1,1,36
1,2,18
1,3,12
1,4,9
1,6,6
2,2,9
2,3,6
3,3,4

How long does it usually take for the AMC results to be released?

Author: admin  //  Category: maths competitions

This AMC was held on the 6th Thursday 2009. This is the Australian Maths Competition, not the American. I know the certificates usually take a month or two, but I recall knowing the results for the prizes and medals within a week.

Well I’ve been a maths teacher in NSW for over 20 years, and we have never received the results within a week. Not sure about the actual date, but your month or two sounds about right.

how can i find books related to kangaroo competition ?

Author: admin  //  Category: maths competitions

Plz tell me the source where i can find books and study material related to the kangaroo math competition .

http://mathkangaroo.org/

What is the mathcounts competition like TEN POINTS ?

Author: admin  //  Category: maths competitions

This is my first time entering mathcounts competition and I wanted to know what it was like. Mathcounts is a math competition. These were my questions:
Can you tell me how everything goes in the mathcounts ? DEtailed answer please !!!!!!
Does everyone in the team have to particiapate?
If they do , then when they take the tests how long does it take for the judges to see if teh snswer is correct?

check out this link it has all the answers to your questions:

http://en.wikipedia.org/wiki/MathCounts

What is domething I can do to beat a cocky genius?

Author: admin  //  Category: maths competitions

There is this genius that is in 4 of my classes. He is literally a natural born genius. I am one of the smartest kids in my grade, but I am not a genius. Anyway, this guy is always cocky and bragging about how he wins all these things because of his genius abilities. What is something I can just rub in his face that I am better at intellectually? I am no good at sports, by the way. I can’t beat him at chess, math competitions, jepordy like games, or anything! Any advice?

Learn to spell "something"?

Geniuses sometimes have problems with simple things and/or common sense. Try playing Simon says with him or a game of Fish.

How to prepare for AMC(australian math competition)?

Author: admin  //  Category: maths competitions

I am going go to math competition this year, and I have no idea how to prepare and score good marks! please help and give an advise.

Hi Mary:),

I’m an online tutor and a college student. One of my specialization is in math competitions, though I focus on ones in United States.

I am, however, familiar with AMC (Australian Math Competition; not to be confused in American Mathematics Competition). The only problem is that I have not seen too many sample problems.

Basically, though, you should review your algebra skills, geometry, probability/counting, and some of the number theory. I am not sure if your AMC tests trigonometry or complex numbers, but review those too.

Lastly, do some actual practices.

If you need further help, feel free to contact me. :)

Good luck!

Where is the William Lowell Putnam mathematics competition held?

Author: admin  //  Category: maths competitions

I am interested in participating in the William L. Putnam maths competition, though I am unsure where it is held. Anyone who could enlighten me on this point would certainly gain a high degree of appreciation. Thanks.

"There are no boundaries."
- Kris Allen, American Idol 2009

Where can i find math problems similar to the ones in the Bergen Academy Math Competition?

Author: admin  //  Category: maths competitions

here’s the website for the BA Math Competition:
http://sites.bergen.org/mathcompetition/exams.asp
i did all of them online, but where can i find more??

Some information is below./

Maths Problems?

Author: admin  //  Category: maths competitions

Hi, I am currently preparing for a maths competitions coming up soon. I received a set of questions to help prepare for it. I can’t solve some of them and it’ll be great you someone could help me out. Working with your solution would be great. Thanks!
1)100 people are standing in a line and they are required to count off in fives as: ‘1,2,3,4,5,1,2,3,4,5,…’ and so on from the 1st person in the line. Anyone who counts ‘5′ must leave the lin. What was the original position in the line of the last person to leave?
2)in the tens digit of a perfect square number is 7, how many units digits are possible. (reasoning please)!
3)When three spherical balls, each of radius 10cm, are placed in a hemispherical dish, it is noticed that the tops of the balls are all exactly level with the top of the dish. What is the radius, in centimetres, of the dish.
4)How MANY ways can you arrange the numbers 1,2,3,4,5,6,7,8,9 in that order with only addition signs. Eg. 1+2+3+45+6+78+9=14
Sorry, for the 4th question, i left out the fact that the numbers must sum up to 144.
clearing up confusion in the 2nd question: after one round of counting to 5, it starts from the beginning again. e.g. once it goes to person 100, it returns to person 1 and the counting begins again.

1)
I had the 47th number left at the end when I used a spreadsheet to actually do this. When I get home this afternoon I will have a think about a way of solving this without actually doing it. If I come up with something I will let you know. Because as this stands I have tried all sorts and I can’t work out how to do it if there were more numbers! Doing it with 100 numbers was labourious enough! (That’s laborious for you americans). My initial thoughts were using modular arithmetic but I can’t find a way for that to work.
It did occur to me that this would be much easier if it were multi choice as it is easier to watch a numbers change in position and work out if it gets deleted than work out which ones stay in. (as I have done below with the 98th).

A few people have said 98 so i will show that to be incorrect.
on the first run through we discard 20 people leaving 80, person 100 is out.
on the second run through we discard 16 people leaving 64 people, and person 99 is gone.
At this point person 98 is a 4 and person 1 is a 5 (and discarded) we then dicard a further 11(thats 12 including person 1) (last person to discard is 61st on that list)
62nd is a 1 63rd is a 2 and 64 is a 3 making the new number 1 a 4
now as we just deleted 13 people we are left with 51 people at this point
we discard person 2, 7, 12, … up to and including the 47th on the list making the 51st now a 4 and the new number 1 a 5 (leaving 37 people)
Next round we get rid of 1, 6, 11, 16, .. 36 (8 people leaving 29) the 37th is a 1, the new 1st is a 2
Next round get rid of 4 , 9, 14, 19, 24 and 29 as 29 is the last on the list this is the original number 98 gone. It is irrelavant that 29 is not divisable by 5

2)
Just one possibility
the ones digit must be a 6
manjyomesando1 has explained this one fairly well already
any number can be written as 10x + y where y is a digit from 0 to 9 and x is any sized number
(10x + y)^2 = 100x^2 + 20xy + y^2
now the 100x^2 is irrelavant, it wont effect the last 2 digits
so 20xy + y^2 is what we need to look at
The ones digit is the last digit of y^2 which is 0, 1, 4, 5, 6, or 9
If y = 0 then 20xy + y^2 = 0 so the 10s digit is 0
if y = 1 then 20xy + y^2 = 20x + 1 for the 10s digit to be 7 is impossible because no matter what x is multiples of 20 will never have a tens digit of 7

if y = 2 then 20xy + y^2 = 40x + 4 again for the 10s digit to be 7 is impossible because no matter what x is multiples of 40 will never have a tens digit of 7

if y = 3 then 20xy + y^2 = 60x + 9 again for the 10s digit to be 7 is impossible because no matter what x is multiples of 60 will never have a tens digit of 7

if y = 4 then 20xy + y^2 = 80x + 9 again for the 10s digit to be 7 is impossible because no matter what x is multiples of 80 will never have a tens digit of 7

if y = 5 then 20xy + y^2 = 100x + 25 This always has a tens digit of 2 (not 7)

if y = 6 then 20xy + y^2 = 120x + 36 now if x = 2 this is 240 + 36 = 276 so 6 is possible

if y = 7 then 20xy + y^2 = 140x + 49 because y^2 is 49 we need to consider if the 10s digit of 140x will ever be a 3 (which it never will because multiples of 14 are even)

if y = 8 then 20xy + y^2 = 160x + 64 so we need to consider if the 10s digit of 160x will ever be a 1 but it never will because m,ultiples of 16 are even

if y = 9 then 20xy + y^2 = 120x + 81 consider if 12x ever ends in a 9, but never because 12x is always even

So we only found one possibility and that is a ones digit of 6

3)
Sketch at:
http://farm4.static.flickr.com/3032/2697142397_80b6bee5dd_o.jpg

Notice the red lines, the long red line is the radius of the dish and it bisects the 60degree angle of the equilateral triangle joining the centre of the spheres and then I have dropped a red line down perpendicular to the equilateral triangle

now from centre of dish to centre of a sphere lets call that d for distance

Cos(30) = r/d
d = r/cos(30)
d = 10/cos(30)
we add this to r (10cm) and we have the radius of the dish

radius of dish = 10 + 10/cos(30)
= 10 + 20/sqrt3
= 21.547cm (3dp)

NOTE: if it were 3r = 30cm then we would be able to fit a whole extra sphere between the 3 that are there but I think it’s obvious this would not fit. The answer must be greater than 20 and less than 30!

4)
someone posted this same question a few days ago, I think it was only 3 (no, it seems it was 4, manjyomesando1 caught one that I originally missed and have since added to my answer, I claim no credit for this extra one - I missed it)

At most you can pair 2 digits together unless it starts with 12, or 13
if it starts with 123 its still too high so at most you can pair 2 digits

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 so a minimum pairing has to be 99 but that’s impossible so we must have at least 2 pairings.

Start by pairing 8 and 9 and see how much short it is
1 + 2 + 3 + 4 + 5 + 6 + 7 + 89 = 117 so we are short by 27 by pairing 3 and 4 we get 27 more so
1 + 2 + 34 + 5 + 6 + 7 + 89 = 144 ******** (1)

Now start by pairing 78 and see how short we are:
1 + 2 + 3 + 4 + 5 + 6 + 78 + 9 = 108 this is 36 short
1 + 2 + 3 + 45 + 6 + 78 + 9 = 144 ********* (2)
12 + 34 + 5 + 6 + 78 + 9 = 144 ************ (3)

Now start by pairing 67 and see how short we are:
1 + 2 + 3 + 4 + 5 + 67 + 8 + 9 = 99 (short by 45 but no single pairing will fix this so we need to try 2 pairings)
12 + 3 + 45 + 67 + 8 + 9 = 144 works ************ (4)

Now is there anythign if we start by pairing 56
1 + 2 + 3 + 4 + 56 + 7 + 8 + 9 = 90, short by 54, would need 2 pairings but 12 + 34 + 56 + 7 + 8 + 9 is only 126 so no good.

1 + 2 + 34 + 5 + 6 + 7 + 89 = 144 ******** (1)
1 + 2 + 3 + 45 + 6 + 78 + 9 = 144 ********* (2)
12 + 34 + 5 + 6 + 78 + 9 = 144 ************ (3)
12 + 3 + 45 + 67 + 8 + 9 = 144 *********** (4)
are the only possibilities