MAA-BIH ORG

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

I’m for a math project about how students experiencing difficulties in Mathematics. Can the educational environment has a problem with it? Or is it the facility has something missing for studnets? Is there a problem within the classroom? Is there any reason how students became difficult to study Math?

If you are suggesting that a student suffers when the learning environment is not conducive to learning, then absolutely, you are correct. Math is difficult for many students. There seems to be a love or hate relationship with it. Those who love it, do well, and those who hate it either don’t do well, or they respect it and try to do as well as they can. I am all for projects if they contribute to the understanding of the class, whether it is math or some other subject.

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./

A king had three sons. After his death, the land and elephants were to be divided as under:
First son = 1/2 share of whole property and elephants
Second son - 1/3rd share
Third son = 1/9th share
The land was divided. But the elephants were 17 in number
which were difficult to be divided among them as per share.
An elephant rider with his elephant passed by. He was told about the problem. He sacrificed and added his own elephant, making the total of 18 elephants. Now division was easy. First son got 1/2 of 18 = 9
Second son got 1/3 of 18 = 6
Third son got 1/9 of 18 = 2
This total was 9+6+2 = 17
The 18th elephant became spare.
The elephant rider took it, rode on it and went away
question is that if rider did’nt come, they were to cut elephants and First son was to get 1/2 of 17 less than 9
Second son 1/3 of 17 less than 6 The third less than 2
but they got full 9 , 6 and 2 by addition of the riders elephant
yet the riders elephant became spare. Is your mathematics ok?

1/2 of 17 = 8.50 = 9
1/3 of 17 = 5.67 = 6
1/9 of 17 = 1.90 = 2

9 + 6 + 2 = 17

Yes, my mathematics are fine. No spare is needed, just use rounding.

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

My wife and I are planning to work part time so that we can take care of our children without using child care. Our background is very scientific (mathematics,biology,physics, etc); we were planning to have a small child care (probably for kids between 5 to 10 years old) in our house that promotes science discovery and appreciation. For example, we would apply the math they learn in school to real problems and explain how more difficult mathematics is used today. But we are not sure how much parents would pay and if they would even be interested.
"Jobs are being lost. People are having a difficult time feeding the family."
That is why parents should invest in kids education.

"I don’t think they would pay much for a babysitter."
I was not talking about a babysitter.

"There is a program that offers something similar…"
Yes, that is what I was talking about.

There is a program that offers something similar, though it comes to you, and isn’t actually a baby sitting service.

Their pricing might help you figure out what you should charge, though.

www.MadScience.com

It would be a very good idea to check with the Better Business Bureau in your area to find out about the average price for local daycare services.

I need to prepare for competition exams like Maths Olympiad, please let me know where can I find papers for the practice of these exams

Hi Sachin S!

According to http://wiki.answers.com/Q/How_do_I_get_Sample_papers_for_maths_olympiad_class_4:

You can print (or solve online) practice paper for Maths Olympiad and CBSE class 4, 5 and 6 at following site:

http://www.examhelp.in

This site generates random questions, so you can generate as many papers you want without questions being repeated.

Here’s wishing you much success!

hello everyone! i have been an extreme screw-up thus far this semester in my analysis class. this unit is mainly focusing on limits (actually not too sure about this I just know it’s an aspect of it and it makes sense b/c the unit is called "LM"). anyways, since i have not completed one assignment, i am not entirely sure of what a limit exactly is. i am good with exponents and numbers in general so i am sure i could piece out the concept pretty quickly. my math teacher is aware that i haven’t done any of the hw, and has just recommended me to a psychologist because he thinks i have extreme focusing problems but my gifted-ness has been masking it. so he told me it didn’t matter how i did on the test i’ll be taking tomorrow about this unit…i would like to do good anyways

my question is…where to start? can someone give me a brief overview of what it is? i know has something (or quite possibly everything) with where functions sometimes converge on graphs. it’s OK to refer me to links, just as long as you’ve actually looked at the sites. i also would like to know the extent of how difficult or complex problems with limits can get. i have an extremely weird math book which basically does the pacing for you; it makes it impossible to just open the book and do a problem without doing the previous ones (which is great for some people). and by complexity of the problems i also mean the type of variety limit problems come in. if that even applies to the topic

THANK YOU
my bad. the first second doesn’t make sense to me but mainly because i’m not even far enough in calculus to be using the greek alphabet on a basis. all i know about limits, is that you have a variable approaching a number. you find what value a function has as x approaches that number. i just think that’s too simple, it’s just algebra really. i would assume my class must be at least one step further than that. i’m sorry if this isn’t making nay sense
woops…**sentence

Well… that’s rather general… I don’t even know what level of calculus you’re in.

Does the first sentence in this article make sense to you?
http://mathworld.wolfram.com/Limit.html

——

Okay, it sounds like you’re in what my school would call either "business calculus" or "engineering calculus". "Analysis" is the name we give to an upper-division math class on the theory of calculus.

Also, believe it or not, but that first sentence is how the limit is actually defined. The notation you’ve seen is actually shorthand for that technical definition, so it’s not quite just algebra - it’s actually much stronger.

Almost all calculus is is evaluating different limits. Of course, we don’t call them limits when they’re special… instead we use terms like "derivative" or "integral", but it all comes down to a limit in the end… so to answer your question on the complexity of limits, it can become quite complex (though to become truly complex, you probably would have to take several more classes).

Maths is my favorite subject.I’m very good at it and always achieve best marks at school.So I would like to find some way or competition that would help me to excel in this field.

AMC, mandelbrot, CAML(for CA residents), ARML, list goes on and on. Look at artofproblemsolving.com for lists of competitions. Don’t worry, there are more than enough,

Suppose the number of students in a mathematics class is x. The teacher insists that each student participate in group work each class. The number of possible groups is given by the following formula.

G= x^2-3x+2 / 2 … the class has 16 How many possible groups are there?

Well given the formula we simply "plug in" 16 that is if x is the number of students we place 16 everywhere see x so:

(16^2-3*(16)+2)/2= 105 combinations.