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Math Problems
 Dear Group,
I couldn't connect with the group for my exam.However,today I am posting some questions.I found them pretty much interesting.I hope you will find them interesting too.
1.A series is formed in the following manner:
A(1) = 1;
A(n) = f(m) numbers of f(m) followed by f(m) numbers of 0;
m is the number of digits in A(n1)
Find A(30). Here f(m) is the remainder when m is divided by 9.
2.Triangle ABC is right angled at B. The bisector of BAC meets BC at D. Let G denote the
centroid (common point of the medians) of the triangle ABC. Suppose that GD is parallel to
AB. Find C.
And another one I will give you.Please don't laugh seeing the question.I wanna be sure if it's answer is 10.
3.There are 20 people in a party excluding you. It is known that you know the same number
of people as you don't know. How many of them do you know?
Let me know what is your answer.Post interesting questions.
Best Wishes
Rafi Post your approaches first!  On Sun, 5/2/10, Abdul Muntakim Rafi <abdulmuntakim_rafi@...> wrote:
From: Abdul Muntakim Rafi <abdulmuntakim_rafi@...>
Subject: [math_club] Math Problems
To: math_club@yahoogroups.com
Date: Sunday, May 2, 2010, 10:02 AMDear Group,
I couldn't connect with the group for my exam.However, today I am posting some questions.I found them pretty much interesting. I hope you will find them interesting too.
1.A series is formed in the following manner:
A(1) = 1;
A(n) = f(m) numbers of f(m) followed by f(m) numbers of 0;
m is the number of digits in A(n1)
Find A(30). Here f(m) is the remainder when m is divided by 9.
2.Triangle ABC is right angled at B. The bisector of BAC meets BC at D. Let G denote the
centroid (common point of the medians) of the triangle ABC. Suppose that GD is parallel to
AB. Find C.
And another one I will give you.Please don't laugh seeing the question.I wanna be sure if it's answer is 10.
3.There are 20 people in a party excluding you. It is known that you know the same number
of people as you don't know. How many of them do you know?
Let me know what is your answer.Post interesting questions.
Best Wishes
RafiHi Rafi
My answer to first question is:
A(30) = A(24) = A(18) = A(12) = A(6) = 77777770000000.
Please confirm.
 In math_club@yahoogroups.com, Abdul Muntakim Rafi <abdulmuntakim_rafi@...> wrote:
>
> Dear Group,
>
> I couldn't connect with the group for my exam.However,today I am posting some questions.I found them pretty much interesting.I hope you will find them interesting too.
>
> 1.A series is formed in the following manner:
> A(1) = 1;
> A(n) =
> f(m) numbers of f(m) followed by f(m) numbers of 0;
> m is the number
> of digits in A(n1)
> Find A(30). Here f(m) is the remainder when m is
> divided by 9.
>
> 2.Triangle ABC is right angled at B. The bisector
> of BAC meets BC at D. Let G denote the
> centroid (common point of the
> medians) of the triangle ABC. Suppose that GD is parallel to
> AB. Find C.
>
> And another one I will give you.Please don't laugh seeing the question.I wanna be sure if it's answer is 10.
>
> 3.There are 20
> people in a party excluding you. It is known that you know the same
> number
> of people as you don't know. How many of them do you know?
>
> Let me know what is your answer.Post interesting questions.
>
> Best Wishes
> Rafi
> Dear Kamal,
Your answer is absolutely right.The sequence cycles after each six words.Thus we can find that A(30)=77777770000000
Actually the questions I posted were from National Olympiad 2010.It took me 20 to 30 minutes to solve the problem there.
I did in the following way
A(1)=1
In case of A(2)
m=number of digit of A(21)=A(1)=1
f(m) is the remainder when m is divided by 9=1
Therefore A(2)=10
In case of A(3)
m=2
f(m)=2
A(3)=2200
In case of A(4)
m=4
f(m)=4
A(4)=44440000
In case of A(5)
m=8
f(m)=8
A(5)=8888888800000000
In case of A(6)
m=16
f(m)=7[f(m) is the remainder when m is divided by 9]
A(6)=77777770000000
In case of A(7)
m=14
f(m)=5
A(7)=5555500000
In case of A(8)
m=10
f(m)=1
A(8)=10
Then A(9)=2200 A(10)=....................................
What is the value of A(30)
We have found that these values will come one after another.
from A(2) to A(7)
These six values will come again and again.
Now except A(1),This series recycle after each six terms.
Then A(30) is the 29th one.Divide 29 with 6.5 is the remainder.The value of 5th one would be the same as A(6)=77777770000000
Answer the other questions and post questions to the group.
Best Regards
Rafi
From: kamallohia <kamallohia@...>
To: math_club@yahoogroups.com
Sent: Mon, May 3, 2010 12:39:13 PM
Subject: [math_club] Re: Math Problems
Hi Rafi
My answer to first question is:
A(30) = A(24) = A(18) = A(12) = A(6) = 77777770000000.
Please confirm.
 In math_club@yahoogrou ps.com, Abdul Muntakim Rafi <abdulmuntakim_ rafi@...> wrote:
>
> Dear Group,
>
> I couldn't connect with the group for my exam.However, today I am posting some questions.I found them pretty much interesting. I hope you will find them interesting too.
>
> 1.A series is formed in the following manner:
> A(1) = 1;
> A(n) =
> f(m) numbers of f(m) followed by f(m) numbers of 0;
> m is the number
> of digits in A(n1)
> Find A(30). Here f(m) is the remainder when m is
> divided by 9.
>
> 2.Triangle ABC is right angled at B. The bisector
> of BAC meets BC at D. Let G denote the
> centroid (common point of the
> medians) of the triangle ABC. Suppose that GD is parallel to
> AB. Find C.
>
> And another one I will give you.Please don't laugh seeing the question.I wanna be sure if it's answer is 10.
>
> 3.There are 20
> people in a party excluding you. It is known that you know the same
> number
> of people as you don't know. How many of them do you know?
>
> Let me know what is your answer.Post interesting questions.
>
> Best Wishes
> Rafi
>
Dear Rafi,
I'm very glad to have you back again. I hope you have passed all of your exams.
Now I'm attaching my solution of our old triangle problem. Tell me if you have problems in reading it, since it's a standard PDF file. Sometimes I experienced problems when downloading attached files from the Yahoo email service through Microsoft Internet Explorer. A good choice may be saving the file onto disk before opening it, or even going over to Mozilla Firefox.
I encourage you to take your first steps in the "mathsforfun" activity. Technically, how will it be? Have you considered to start a blog, or are you planning to go Facebook instead?
There you have my contribution. It's a bunch of eight problems of varying difficulty. I hope they will do.
(1) Give an equation from which an eggshaped curve can be plotted.
(2) Three parallel strips have been painted on the floor of an airport lobby. Their lengths are 8, 11, and 13 meters. An architect decided later that the strips should all have the same length. Repainting and/or unpainting the strips cost the same. In order to minimize costs, what length will the strips finally have?
(3) The perimeter of a rightangle triangle is 96. The altitude on the hypothenuse is 19.2. What are the lengths of its sides?
(4) The sum of all positive multiples of three that are less than 1000 is multiplied by six and added to four times the sum of the positive nonmultiples of three that are less than 1000. What is the resulting sum?
(5) For a given arithmetic series, the sum of the first 50 terms is 200 and the sum of the next 50 terms is 2700. What is the first term in the series?
(6) Does equation 1 / (x  1) = x / (x^2  1) have any solution?
(7) What are the real solutions of x = sqrt(2 * x + 3)?
(8) A fivelegged Martian has a drawer full of socks, each of which is red, white, or blue, and there are at least five socks of each color. The Martian pulls out one sock at a time without looking. What is the least number of socks that the Martian must remove from the drawer to be certain there will be 5 socks of the same color?
That's all by now.
Best regards,
Jon Tavasanis Dear Group,
We opened a Science Club in our school.I am taking the Maths classes there.I have taken only one class.Can you give me any suggestion on what should I discuss with the students of 8 and 9.I am in class 10.Which questions can ensure the increase of their ability of doing maths.Tomorrow I will discuss with them about the basic knowledge of maths.Can advise me anything more?
Best Wishes
Rafi  Hi,
It is really great to hear that you have started a science club in your school.
I wrote some notes and guidelines on BdMO. This is available in the website of Kushtia Math Circle.
Just check in the document section.
Regards
Moon
From: "math_club@yahoogroups.com" <math_club@yahoogroups.com>
To: math_club@yahoogroups.com
Sent: Wednesday, May 19, 2010 20:10:31
Subject: [math_club] Digest Number 1408math_club Messages In This Digest (1 Message)
 1a.
 Math Problems From: Abdul Muntakim Rafi
Message
 1a.
Math Problems
Posted by: "Abdul Muntakim Rafi" abdulmuntakim_rafi@... abdulmuntakim_rafi
Tue May 18, 2010 7:58 am (PDT)
Dear Group,
We opened a Science Club in our school.I am taking the Maths classes there.I have taken only one class.Can you give me any suggestion on what should I discuss with the students of 8 and 9.I am in class 10.Which questions can ensure the increase of their ability of doing maths.Tomorrow I will discuss with them about the basic knowledge of maths.Can advise me anything more?
Best Wishes
Rafi
Need to Reply?Click one of the "Reply" links to respond to a specific message in the Daily Digest.
MARKETPLACE
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Visit Your Group  Yahoo! Groups Terms of Use  Unsubscribe  Dear Moon,
Thanks for the response.I have downloaded almost all document from your site.But I think at first the students should build up their basic in maths.So we will discuss with them on the basic rules of maths.Then I will try to give them the pages for studying.We can solve the problems in the Science Club.
Best Wishes
Rafi
From 1 to 1000, how many integers are multiples of 3 or 6 but not of 5?
You wrote that the answer is 378.Is it right?
From: Tarik Adnan <moonmath420@...>
To: math_club@yahoogroups.com
Sent: Thu, May 20, 2010 5:24:57 PM
Subject: [math_club] Re: Math Problems
Hi,
It is really great to hear that you have started a science club in your school.
I wrote some notes and guidelines on BdMO. This is available in the website of Kushtia Math Circle.
Just check in the document section.
Regards
Moon
From: "math_club@yahoogro ups.com" <math_club@yahoogrou ps.com>
To: math_club@yahoogrou ps.com
Sent: Wednesday, May 19, 2010 20:10:31
Subject: [math_club] Digest Number 1408math_club Messages In This Digest (1 Message)
 1a.
 Math Problems From: Abdul Muntakim Rafi
Message
 1a.
Math Problems
Posted by: "Abdul Muntakim Rafi" abdulmuntakim_ rafi@yahoo. com abdulmuntakim_ rafi
Tue May 18, 2010 7:58 am (PDT)
Dear Group,
We opened a Science Club in our school.I am taking the Maths classes there.I have taken only one class.Can you give me any suggestion on what should I discuss with the students of 8 and 9.I am in class 10.Which questions can ensure the increase of their ability of doing maths.Tomorrow I will discuss with them about the basic knowledge of maths.Can advise me anything more?
Best Wishes
Rafi
Need to Reply?Click one of the "Reply" links to respond to a specific message in the Daily Digest.
MARKETPLACE
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Visit Your Group  Yahoo! Groups Terms of Use  Unsubscribe  Hello Group,
Here are some problems for you.
1.At a party, there girls and three boys sit at a round table. What
is the probability that no two persons of the same sex sit at
adjacent places?
2.The base 10 representation of 6^4 is 1296 . What is the base6
representation of 1295 ?
There is an interesting thing about the second problem.I didn't know anything about base representation of anything before.But I solved the problem.Please tell me how you solve it.
Best Wishes
Rafi  Dear Rafi:When teaching mathematics, I always remind myself what Sofia Kovalevskaya once said: "It is impossible to be a mathematician without being a poet in soul." Therefore, the first thing I suggest you to do is make your class interactive and poetic, not prosaic like 2 × 3 = 6. Tell your students story, tell them history, tell them about great persons like Omar Khyaam, who was one of World's finest poets and mathematicians. And construct your problems and solutions in storylines, so mathematics would be lively and concrete to them, rather than abstract and formidable.About chosing your lecture contents, you could wish to consult this website: http://www.aaamath.com/grade8.htmlWish you good luck!Kind regards,S. M. Mamun Ar Rashid
 On Tue, 5/18/10, Abdul Muntakim Rafi <abdulmuntakim_rafi@...> wrote:
From: Abdul Muntakim Rafi <abdulmuntakim_rafi@...>
Subject: [math_club] Math Problems
To: math_club@yahoogroups.com
Date: Tuesday, May 18, 2010, 7:58 AMDear Group,
We opened a Science Club in our school.I am taking the Maths classes there.I have taken only one class.Can you give me any suggestion on what should I discuss with the students of 8 and 9.I am in class 10.Which questions can ensure the increase of their ability of doing maths.Tomorrow I will discuss with them about the basic knowledge of maths.Can advise me anything more?
Best Wishes
Rafi  You can also browse through the archive: http://mathforum.org/dr/math/
It will surely help to prepare for classes.
regards,
Subeen.  (1) Problem type: Circular Permutation, ProbabilityIn a circular arrangement of people, you must keep one person fixed, and then work on arranging the remaining people. This is because 123 & 231 are different permutations in line arrangement, but the same permutation in circular/round arrangement. So, when one person is fixed in round arrangement, any change in arrangements of the rest will create a new permutation. This way, duplicity is avoided.Therefore, let's fix a girl in a position. Now we have 2 other girls and 3 boys to arrange. 5 people can be arranged in total 5! or (5.4.3.2.1) or 120 ways.But if we want to keep boys and girls in alternate positions, first we need to place one group, and then place the other group in gaps of the first group. So, after fixing one girl in her position, the 2 other girls can be placed in 2! ways and the 3 boys can be placed in 3! ways in the girls' gaps. Therefore, total no. of ways here is 2! × 3! or (2.1) × (3.2.1) or 12 ways.Probability = 12/120 or 1/10.Note: There is another issue. Some problems might specify that clockwise and anticlockwise arrangements are to be considered the same. In such case, you would divide the total no. of arrangements by 2.(2) Problem type: Number System, Base, Conversion Of BaseSuppose, you have bought an icecream the price of which is 31 Bangladeshi Taka (BDT). How will you pay the seller the money? Perhaps you would give her one 1BDT note and three 10BDT notes? Why? Because we know that 31 amounts to one 1 plus three 10s.But if you lived in ancient Babylon (presentday Iraq) and bought the ice cream there for 31 Babylonian Taka (BBT), then you would pay the seller one 1BBT note but three 60BBT notes. Why is the difference between modern Bangladesh and ancient Babylon? Because in Bangladesh, and all most all over the world today, we use 10base number system, but the ancient Babylonians used 60base number system.The scenario would be yet different in ancient farway Maya Kingdom in Central America that used 20base number system. For 31, the May would pay one 1Maya Taka note but three 20Maya Taka notes.Although it may sound puzzling, the matter is simple. In our 10base system,5327 (decimal or 10base)= (7×1) + (2×10) + (3×100) + (5×1000)= (7×10^0) + (2×10^1) + (3×10^2) + (5×10^3)In ancient Maya system,5327 (Maya)= (7×20^0) + (2×20^1) + (3×20^2) + (5×20^3)= 7 + 40 + 1,200 + 40,000 (now converted into decimal)= 41,247 (in decimal)In 6base system,5327 (base 6)= (7×6^0) + (2×6^1) + (3×6^2) + (5×6^3)= 7 + 12 + 108 + 1080 (now converted into decimal)= 1207 (in decimal)So, here we are going from a particular base to decimal. We can also convert a decimal number to particular base number. To do this quickly, first we need to arrange powers of that base, and then chose suitable combinations from those powers (starting from the highest) to form the desired number:So, what is the base6 representation of 1295 (which is in base10, i.e. decimal)?Step 1: Arrange “Powers of 6”:6^0, 6^1, 6^2, 6^3, 6^4…=1, 6, 36, 216, 1296…Step 2: Choose suitable combinations of these powers 6, 36, 216, 1296… to form 1295.We cannot choose 1296; it is beyond limit. So, chose the next power as many times as possible. We must choose 216 five times.5×216 (=1080)We now have 12951080=215 remaining. Go to 36 maximum possible, i.e., five times.5×36 (=180)Now we have 215180=35. Go next to 6 five times: 5×6 (=30). Finally, we have 5.Therefore, 1295 (in decimal)=1080+180+30+5= (5×216) + (5×36) + (5×6) + (5×1)= (5×6^3) + (5×6^2) + (5×6^1) + (5×6^0)= 5555 (in base6)Regards,S. M. Mamun Ar Rashid
 On Thu, 5/20/10, Abdul Muntakim Rafi <abdulmuntakim_rafi@...> wrote:
From: Abdul Muntakim Rafi <abdulmuntakim_rafi@...>
Subject: [math_club] Math Problems
To: math_club@yahoogroups.com
Date: Thursday, May 20, 2010, 9:34 AMHello Group,
Here are some problems for you.
1.At a party, there girls and three boys sit at a round table. What
is the probability that no two persons of the same sex sit at
adjacent places?
2.The base 10 representation of 6^4 is 1296 . What is the base6
representation of 1295 ?
There is an interesting thing about the second problem.I didn't know anything about base representation of anything before.But I solved the problem.Please tell me how you solve it.
Best Wishes
Rafi  From 1 to 1000, how many integers are multiples of 3 or 6 but not of 5?
You wrote that the answer is 378.Is it right?My Answer to your question is 267.
From: Abdul Muntakim Rafi <abdulmuntakim_rafi@...>
To: math_club@yahoogroups.com
Sent: Thu, May 20, 2010 8:41:08 PM
Subject: Re: [math_club] Re: Math Problems
Dear Moon,
Thanks for the response.I have downloaded almost all document from your site.But I think at first the students should build up their basic in maths.So we will discuss with them on the basic rules of maths.Then I will try to give them the pages for studying.We can solve the problems in the Science Club.
Best Wishes
Rafi
From 1 to 1000, how many integers are multiples of 3 or 6 but not of 5?
You wrote that the answer is 378.Is it right?
From: Tarik Adnan <moonmath420@ yahoo.com>
To: math_club@yahoogrou ps.com
Sent: Thu, May 20, 2010 5:24:57 PM
Subject: [math_club] Re: Math Problems
Hi,
It is really great to hear that you have started a science club in your school.
I wrote some notes and guidelines on BdMO. This is available in the website of Kushtia Math Circle.
Just check in the document section.
Regards
Moon
From: "math_club@yahoogro ups.com" <math_club@yahoogrou ps.com>
To: math_club@yahoogrou ps.com
Sent: Wednesday, May 19, 2010 20:10:31
Subject: [math_club] Digest Number 1408Messages In This Digest (1 Message)
 1a.
 Math Problems From: Abdul Muntakim Rafi
Message
 1a.
Math Problems
Posted by: "Abdul Muntakim Rafi" abdulmuntakim_ rafi@yahoo. com abdulmuntakim_ rafi
Tue May 18, 2010 7:58 am (PDT)
Dear Group,
We opened a Science Club in our school.I am taking the Maths classes there.I have taken only one class.Can you give me any suggestion on what should I discuss with the students of 8 and 9.I am in class 10.Which questions can ensure the increase of their ability of doing maths.Tomorrow I will discuss with them about the basic knowledge of maths.Can advise me anything more?
Best Wishes
Rafi
Need to Reply?Click one of the "Reply" links to respond to a specific message in the Daily Digest.
MARKETPLACE
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Visit Your Group  Yahoo! Groups Terms of Use  Unsubscribe  Dear Kamal,
My answer is also 267.It was a question from divisional Olympiad 2010.I found it in KMC's site.There the given answer was 378.It was a mistake.I wrote to Moon about it so that he could see the mistake.He is the administrator of KMC.However,thanks for the response.
Best Wishes
Rafi
From: kamal lohia <kamallohia@...>
To: math_club@yahoogroups.com
Cc: Abdul Muntakim Rafi <abdulmuntakim_rafi@...>
Sent: Fri, May 21, 2010 3:55:45 PM
Subject: Re: [math_club] Re: Math Problems
From 1 to 1000, how many integers are multiples of 3 or 6 but not of 5?
You wrote that the answer is 378.Is it right?My Answer to your question is 267.
From: Abdul Muntakim Rafi <abdulmuntakim_ rafi@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Thu, May 20, 2010 8:41:08 PM
Subject: Re: [math_club] Re: Math Problems
Dear Moon,
Thanks for the response.I have downloaded almost all document from your site.But I think at first the students should build up their basic in maths.So we will discuss with them on the basic rules of maths.Then I will try to give them the pages for studying.We can solve the problems in the Science Club.
Best Wishes
Rafi
From 1 to 1000, how many integers are multiples of 3 or 6 but not of 5?
You wrote that the answer is 378.Is it right?
From: Tarik Adnan <moonmath420@ yahoo.com>
To: math_club@yahoogrou ps.com
Sent: Thu, May 20, 2010 5:24:57 PM
Subject: [math_club] Re: Math Problems
Hi,
It is really great to hear that you have started a science club in your school.
I wrote some notes and guidelines on BdMO. This is available in the website of Kushtia Math Circle.
Just check in the document section.
Regards
Moon
From: "math_club@yahoogro ups.com" <math_club@yahoogrou ps.com>
To: math_club@yahoogrou ps.com
Sent: Wednesday, May 19, 2010 20:10:31
Subject: [math_club] Digest Number 1408Messages In This Digest (1 Message)
 1a.
 Math Problems From: Abdul Muntakim Rafi
Message
 1a.
Math Problems
Posted by: "Abdul Muntakim Rafi" abdulmuntakim_ rafi@yahoo. com abdulmuntakim_ rafi
Tue May 18, 2010 7:58 am (PDT)
Dear Group,
We opened a Science Club in our school.I am taking the Maths classes there.I have taken only one class.Can you give me any suggestion on what should I discuss with the students of 8 and 9.I am in class 10.Which questions can ensure the increase of their ability of doing maths.Tomorrow I will discuss with them about the basic knowledge of maths.Can advise me anything more?
Best Wishes
Rafi
Need to Reply?Click one of the "Reply" links to respond to a specific message in the Daily Digest.
MARKETPLACE
Change settings via the Web (Yahoo! ID required)
Change settings via email: Switch delivery to Individual  Switch format to Traditional
Visit Your Group  Yahoo! Groups Terms of Use  Unsubscribe  Dear Group,
Here are some problems for u.They aren't very hard.But you will pass your time well with them because they are very interesting.They seemed very interesting to me.You don't need to apply anything here.It's a lot like puzzle.
1.Five men are wearing either blue or red hats. A man wearing a blue hat always tells
the truth. A man wearing a red hat always lies. They look at each other’s hat.
A Says “I see 4 blue hats.”
B Says “I see 3 blue hats and 1 red hat.”
C Says “I see 1 blue hat and 3 red hats.”
D says “I see 4 red hats.”
E Does not say anything.
Find the color of each person’s hat.
2.Asmaa, and her brother Ahmed are chess players. Asmaa's son Shamim and her daughter Sharmeen are also chess players. The worst player's twin (who
is one of the 4 chess players) and best player are of the opposite sex. The
worst player and the best player are the same age. Who is the worst player?
Best Wishes
Rafi  Dear Group,
Here are some problems.They are not easy.
1.Find all the prime numbers p and positive integers a and b such that p^a + p^b is the square of an integer.
2.In a set of 131 natural numbers, no number has a prime factor greater than 42. Prove that it is possible to choose four numbers from this set such that their product is a perfect square.
If you can't solve it,please discuss it with other members of the group.
I have some questions.In 10th question does the 131 natural numbers have any relation with the problem.if it has,what it is?
I will be waiting for ur answer.
Best Wishes
Rafi  Hi Rafi
1) if a = b, then p^a + p^b = 2*p^a .
Now p must be 2 only and a can be any odd positive integer.
If a < b, then p^a + p^b = p^a(1 + p^ba) can never be perfect square.
From: Abdul Muntakim Rafi <abdulmuntakim_rafi@...>
To: math_club@yahoogroups.com
Sent: Mon, June 7, 2010 12:13:02 AM
Subject: [math_club] Math Problems
Dear Group,
Here are some problems.They are not easy.
1.Find all the prime numbers p and positive integers a and b such that p^a + p^b is the square of an integer.
2.In a set of 131 natural numbers, no number has a prime factor greater than 42. Prove that it is possible to choose four numbers from this set such that their product is a perfect square.
If you can't solve it,please discuss it with other members of the group.
I have some questions.In 10th question does the 131 natural numbers have any relation with the problem.if it has,what it is?
I will be waiting for ur answer.
Best Wishes
Rafi
If a < b, then p^a + p^b = p^a(1 + p^ba) can never be perfect square.>>>>>>>
How do you know??? Take a=2k, b=2k+3, p=2
Solution:
When a=b, we get, p=2, a=2k+1
When WLOG a<b
As are relatively prime, they are both squares, i.e. a=2k
Let ba=x (just for convenience, btw x is odd)
So,
Now we consider two cases:
1. p odd
Then ; which implies, m1=1, so
2.p=2
That is
Here , which implies
Problem 2 is a folklore (classic) problem; IMO 1985/4, APMO 2007/1 and many more problems including this one use the same idea.
List the primes less than 42, write the numbers as product of prime powers, and argue with pigeonhole principle.
I mean let the 131 numbers be, and the 13 primes less than 42, , Write
By the way, what is the source of the second problem?
Moon
From: "math_club@yahoogroups.com" <math_club@yahoogroups.com>
To: math_club@yahoogroups.com
Sent: Monday, June 7, 2010 19:39:44
Subject: [math_club] Digest Number 1420math_club Messages In This Digest (2 Messages)
 1a.
 Math Problems From: Abdul Muntakim Rafi
 1b.
 Re: Math Problems From: kamal lohia
Messages
 1a.
Math Problems
Posted by: "Abdul Muntakim Rafi" abdulmuntakim_rafi@... abdulmuntakim_rafi
Sun Jun 6, 2010 11:43 am (PDT)
Dear Group,
Here are some problems.They are not easy.
1.Find all the prime numbers p and positive integers a and b such that p^a + p^b is the square of an integer.
2.In a set of 131 natural numbers, no number has a prime factor greater than 42. Prove that it is possible to choose four numbers from this set such that their product is a perfect square.
If you can't solve it,please discuss it with other members of the group.
I have some questions.In 10th question does the 131 natural numbers have any relation with the problem.if it has,what it is?
I will be waiting for ur answer.
Best Wishes
Rafi
 1b.
Re: Math Problems
Posted by: "kamal lohia" kamallohia@... kamallohia
Sun Jun 6, 2010 10:06 pm (PDT)
Hi Rafi
1) if a = b, then p^a + p^b = 2*p^a .
Now p must be 2 only and a can be any odd positive integer.
____________ _________ _________ __
From: Abdul Muntakim Rafi <abdulmuntakim_ rafi@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Mon, June 7, 2010 12:13:02 AM
Subject: [math_club] Math Problems
Dear Group,
Here are some problems.They are not easy.
1.Find all the prime numbers p and positive integers a and b such that p^a + p^b is the square of an integer.
2.In a set of 131 natural numbers, no number has a prime factor greater than 42. Prove that it is possible to choose four numbers from this set such that their product is a perfect square.
If you can't solve it,please discuss it with other members of the group.
I have some questions.In 10th question does the 131 natural numbers have any relation with the problem.if it has,what it is?
I will be waiting for ur answer.
Best Wishes
Rafi
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MARKETPLACE
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Visit Your Group  Yahoo! Groups Terms of Use  Unsubscribe Dear moon,
Thanks for ur reply.
I posted the problems from the National Olympiad BDMO 2010.The 2nd question was from the Secondary group(10th question) and the 1st question was also from BDMO Secondary and Higher Secondary Group.
Best wishes
Rafi
From: Tarik Adnan <moonmath420@...>
To: math_club@yahoogroups.com
Sent: Mon, June 7, 2010 10:13:16 PM
Subject: [math_club] Re: Math Problems
If a < b, then p^a + p^b = p^a(1 + p^ba) can never be perfect square.>>>>>>>
How do you know??? Take a=2k, b=2k+3, p=2
Solution:
When a=b, we get, p=2, a=2k+1
When WLOG a<b
As are relatively prime, they are both squares, i.e. a=2k
Let ba=x (just for convenience, btw x is odd)
So,
Now we consider two cases:
1. p odd
Then ; which implies, m1=1, so
2.p=2
That is
Here , which implies
Problem 2 is a folklore (classic) problem; IMO 1985/4, APMO 2007/1 and many more problems including this one use the same idea.
List the primes less than 42, write the numbers as product of prime powers, and argue with pigeonhole principle.
I mean let the 131 numbers be, and the 13 primes less than 42, , Write
By the way, what is the source of the second problem?
Moon
From: "math_club@yahoogro ups.com" <math_club@yahoogrou ps.com>
To: math_club@yahoogrou ps.com
Sent: Monday, June 7, 2010 19:39:44
Subject: [math_club] Digest Number 1420math_club Messages In This Digest (2 Messages)
 1a.
 Math Problems From: Abdul Muntakim Rafi
 1b.
 Re: Math Problems From: kamal lohia
Messages
 1a.
Math Problems
Posted by: "Abdul Muntakim Rafi" abdulmuntakim_ rafi@yahoo. com abdulmuntakim_ rafi
Sun Jun 6, 2010 11:43 am (PDT)
Dear Group,
Here are some problems.They are not easy.
1.Find all the prime numbers p and positive integers a and b such that p^a + p^b is the square of an integer.
2.In a set of 131 natural numbers, no number has a prime factor greater than 42. Prove that it is possible to choose four numbers from this set such that their product is a perfect square.
If you can't solve it,please discuss it with other members of the group.
I have some questions.In 10th question does the 131 natural numbers have any relation with the problem.if it has,what it is?
I will be waiting for ur answer.
Best Wishes
Rafi
 1b.
Re: Math Problems
Posted by: "kamal lohia" kamallohia@yahoo. com kamallohia
Sun Jun 6, 2010 10:06 pm (PDT)
Hi Rafi
1) if a = b, then p^a + p^b = 2*p^a .
Now p must be 2 only and a can be any odd positive integer.
____________ _________ _________ __
From: Abdul Muntakim Rafi <abdulmuntakim_ rafi@yahoo. com>
To: math_club@yahoogrou ps.com
Sent: Mon, June 7, 2010 12:13:02 AM
Subject: [math_club] Math Problems
Dear Group,
Here are some problems.They are not easy.
1.Find all the prime numbers p and positive integers a and b such that p^a + p^b is the square of an integer.
2.In a set of 131 natural numbers, no number has a prime factor greater than 42. Prove that it is possible to choose four numbers from this set such that their product is a perfect square.
If you can't solve it,please discuss it with other members of the group.
I have some questions.In 10th question does the 131 natural numbers have any relation with the problem.if it has,what it is?
I will be waiting for ur answer.
Best Wishes
Rafi
Need to Reply?Click one of the "Reply" links to respond to a specific message in the Daily Digest.
MARKETPLACE
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Change settings via email: Switch delivery to Individual  Switch format to Traditional
Visit Your Group  Yahoo! Groups Terms of Use  Unsubscribe  Hmm...the second one was a bit hard for secondary, however, it was the last one...so I guess it is OK...
I did not look at the problem sets after BdMO, and I did not recognize it as I am solving a lot of problem lately :s
btw the first problem you posted was a bit simplified than the original one. We need to find a,b such that p^a+p^b is a square of a RATIONAL number, not just integer. The solution method is same, but a bit more case work and modification is needed.
I could not complete this one in BdMO because I ran out of time...:(...btw I won the contest...so who cares :p
Moon
From: "math_club@yahoogroups.com" <math_club@yahoogroups.com>
To: math_club@yahoogroups.com
Sent: Wednesday, June 9, 2010 19:07:15
Subject: [math_club] Digest Number 1422math_club Messages In This Digest (1 Message)
 1a.
 Re: Math Problems From: Abdul Muntakim Rafi
Message
 1a.
Re: Math Problems
Posted by: "Abdul Muntakim Rafi" abdulmuntakim_rafi@... abdulmuntakim_rafi
Tue Jun 8, 2010 9:26 am (PDT)
Dear moon,
Thanks for ur reply.
I posted the problems from the National Olympiad BDMO 2010.The 2nd question was from the Secondary group(10th question) and the 1st question was also from BDMO Secondary and Higher Secondary Group.
Best wishes
Rafi