Post mathematic related questions here

[rst]

Re: mathematic related questions

take example which I gave in previous post

2 ⁴ +3 ⁴ =97 ( prime)
it is possible for n=4

 

[mastercool8695]

Re: mathematic related questions

ok..
got that now..

do you know the answer for your "whats next in series" one ?

cuz, i cant decipher the relation..

 

[rst]

Re: mathematic related questions

----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION

*img836.imageshack.us/img836/1482/nextno.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------

ans is 55
--------------------------------------------------------------------
Difference between 1st and 3rd number =4(2² )
Difference between 3rd and 5th number =9(3² )
Difference between 5th and 7th number =16(4² )

so Difference between 7th and 9th number =25(5² )

hence number= 30+25
=55

 

[mastercool8695]

Re: mathematic related questions

55 is the 9th number,

anyways,
i was at some time, looking at the even and odd placed numbers together, but couldn't find the link..

 

[rst]

Re: mathematic related questions

ok, edited.

 

[mastercool8695]

Re: mathematic related questions

x[SUP]2[/SUP] + y[SUP]2[/SUP] = 2007
how many solutions (x,y) exist such that x and y are positive integers

A)NONE
B)EXACTLY 2
C)>2 BUT FINITELY MANY
D)INFINITELY MANY

 

[rst]

Re: mathematic related questions

x[SUP]2[/SUP] + y[SUP]2[/SUP] = 2007
how many solutions (x,y) exist such that x and y are positive integers

A)NONE
B)EXACTLY 2
C)>2 BUT FINITELY MANY
D)INFINITELY MANY

Ans is A)NONE
------------------------------------------------------------------------
x ² + y ² = 2007
y=√ (2007-x²)

It is defined for (2007-x² ) ≥ 0
2007 ≥ x²
√ 2007 ≥ x
44.79 ≥ x

As x is positive integer
x ∈ {1,2,..............,44}

these finite values of x will give finite values of y
So option (D) is rejected

As y is also positive integer
So in y=√ (2007-x²)
(2007-x²) should give square number
put finite values of x and check it

(2007-x²) will not give square value for any value of x from 1 to 44
------------------------------------------------------------------------------------
If you find this solution hard
then also there is another( easy) method of solving this question

 

[Niilesh]

Re: mathematic related questions

Other method
------------------------------------------------------------------
aⁿ+bⁿ is prime number
where n is positive integer other than 1
here n can be even or odd

case 1:- If n is odd
aⁿ+bⁿ =prime number
(a+b) (aⁿ⁻¹ - aⁿ⁻² b² +................+(-1)ⁿ⁻¹ bⁿ⁻¹ ) =prime number
Not possible as here prime number is product of two numbers

Spoiler

case 2:- If n is even

I)Let n =2m where m=odd
aⁿ+bⁿ = prime number
a^2m+ b^2m = prime number
(a² )^m+(b² )^m = prime number
so aⁿ+bⁿ can't be prime number (by case 1)

Hence aⁿ+bⁿ = prime number is only possible if n=2m where m is even
i.e n= 4,8,12,16,20,24,………………………………

C) possible n=12,20,24,.......................
They can be expressed as
12=3 x 8
20= 5x 4
24=3 x 8
.
.
similar to case 2(I) not prime number

B)here possible values of n =4,8,16,...................
if aⁿ+bⁿ = prime number
then it will definitely be prime number for these values of n

Can you explain this expansion? - aⁿ+bⁿ = (a+b) (aⁿ⁻¹ - aⁿ⁻² b² +................+(-1)ⁿ⁻¹ bⁿ⁻¹ ) when 'n' is odd

 

[rst]

Re: mathematic related questions

This is an identity
It is valid when n is odd

For example
a³+b³ =(a+b) (a²-ab+b²)

EDIT : aⁿ+bⁿ =(a+b) (aⁿ⁻¹ - aⁿ⁻² b +................+(-1)ⁿ⁻¹ bⁿ⁻¹ )

 

[Niilesh]

Re: mathematic related questions

Ans is A)NONE
------------------------------------------------------------------------
x ² + y ² = 2007
y=√ (2007-x²)

It is defined for (2007-x² ) ≥ 0
2007 ≥ x²
√ 2007 ≥ x
44.79 ≥ x

As x is positive integer
x ∈ {1,2,..............,44}

these finite values of x will give finite values of y
So option (D) is rejected

As y is also positive integer
So in y=√ (2007-x²)
(2007-x²) should give square number
put finite values of x and check it

(2007-x²) will not give square value for any value of x from 1 to 44
------------------------------------------------------------------------------------
If you find this solution hard
then also there is another( easy) method of solving this question

That's a little unconvincing statement
Isn't there any way to prove that statement, except putting all the no.s?



BTW My method :
x² + y² = 2007 = 2004 + 3 = 4K +3 (where k = 2004/4)
Now we know that any perfect square when divided by 4 gives remainder either 0 or 1
So
x² + y² = 4K
or x² + y² = 4K +1
or x² + y² = 4K +2
Hence, no integral values of (x,y) will be possible.

This is an identity
It is valid when n is odd

For example
a³+b³ =(a+b) (a²-ab+b²)

EDIT : aⁿ+bⁿ =(a+b) (aⁿ⁻¹ - aⁿ⁻² b +................+(-1)ⁿ⁻¹ bⁿ⁻¹ )

It is derived from binomial expansion/equation or it was also proved with PMI ?

 

[rst]

Re: mathematic related questions

It is derived from binomial expansion/equation or it was also proved with PMI ?

I used the identity directly (as we do with the formulas)


Anyway we can derive this identity

 

[mastercool8695]

Re: mathematic related questions

^^ @nillesh : couldn't get the point buddy..

 

[rst]

Re: mathematic related questions

BTW My method :
x² + y² = 2007 = 2004 + 3 = 4K +3 (where k = 2004/4)
Now we know that any perfect square when divided by 4 gives remainder either 0 or 1
So
x² + y² = 4K
or x² + y² = 4K +1
or x² + y² = 4K +2
Hence, no integral values of (x,y) will be possible.

I think you mean
----------------------------------
2007 will give remainder 3 when divided by 4
x² + y² will give remainder 0,1,2 when divided by 4
As 2007 give remainder other than 0,1,2 so there is no solution
---------------------------------------------
But if we take example
x² + y² = 102
here 102 give remainder 2 when divided by 4
x² + y² will give remainder 0,1,2 when divided by 4
But even then there is no solution

 

[Niilesh]

Re: mathematic related questions

I think you mean
----------------------------------
2007 will give remainder 3 when divided by 4
x² + y² will give remainder 0,1,2 when divided by 4
As 2007 give remainder other than 0,1,2 so there is no solution
---------------------------------------------
But if we take example
x² + y² = 102
here 102 give remainder 2 when divided by 4
x² + y² will give remainder 0,1,2 when divided by 4
But even then there is no solution

you got it wrong, I didn't said that if the remainder is 0, 1, 2 then we will get solutions. I said that if the remainder is 3 then their will be no solution

^^ @nillesh : couldn't get the point buddy..

It's Niilesh(or Nilesh whatever you like)
I will explain the solution tomorrow when i will be online from PC

 

[rst]

Re: mathematic related questions

I think you mean
----------------------------------
2007 will give remainder 3 when divided by 4
x² + y² will give remainder 0,1,2 when divided by 4
As 2007 give remainder other than 0,1,2 so there is no solution

I also meant the same
bolded line means remainder is 3

 

[Niilesh]

Re: mathematic related questions

I also meant the same
bolded line means remainder is 3

that line is correct
I was replying to your example

 

[rst]

Re: mathematic related questions

that line is correct
I was replying to your example

yeah

If a number number gives remainder 3 when divided by 4
then we have no solution in "x² + y²=number"

we can use this method with number giving remainder 3

But there are other number (as I mentioned in my example) which give no solution in "x² + y²=number"

 

[Niilesh]

Re: mathematic related questions

^^ @nillesh : couldn't get the point buddy..

I will re frame my answer
Note: Any perfect square when divided by 4 gives remainder either 0 or 1
Proof:
We know that all numbers can be in the of form 4k or 4k+1 or 4k+2 or 4k+3 (where K is some integer)
Case 1: No. is of the form 4K
=>(4K)²= 4a + 0 Remainder is 0

Case 2: No. is of the form 4K+1
=>(4K + 1)²= 4b + 1 Remainder is 1

Case 3: No. is of the form 4K+2
=>(4K+2)²= 4c + 2² = 4d Remainder is 0

Case 4: No. is of the form 4K+3
=>(4K+3)² = 4e +3² = 4f +1 Remainder is 1
Hence, Proved.
----------------------------------------------------------------------------------------
Now it is given that
x² + y² = 2007 = 2004 + 3 = 4K +3 (where k = 2004/4)
We know that any perfect square when divided by 4 gives remainder either 0 or 1
So
x² + y² = 4K (if Both are x and y are even)
or x² + y² = 4K +1 (if one is even and other is odd)
or x² + y² = 4K +2 (if Both are odd)

As 2007 give remainder other than 0,1,2 so no integral values of (x,y) will be possible

yeah

If a number number gives remainder 3 when divided by 4
then we have no solution in "x² + y²=number"

we can use this method with number giving remainder 3

But there are other number (as I mentioned in my example) which give no solution in "x² + y²=number"

ya, your method is better but a little lengthy.
BTW if I had to solve such questions i would have probably applied hit and trial using the following conditions :

x² + y² = 4K (if Both are x and y are even)
or x² + y² = 4K +1 (if one is even and other is odd)
or x² + y² = 4K +2 (if Both are odd)

 

[rst]

Re: mathematic related questions

I will re frame my answer
Note: Any perfect square when divided by 4 gives remainder either 0 or 1
Proof:
We know that all numbers can be in the of form 4k or 4k+1 or 4k+2 or 4k+3 (where K is some integer)
Case 1: No. is of the form 4K
=>(4K)²= 4a + 0 Remainder is 0

Case 2: No. is of the form 4K+1
=>(4K + 1)²= 4b + 1 Remainder is 1

Case 3: No. is of the form 4K+2
=>(4K+2)²= 4c + 2² = 4d Remainder is 0

Case 4: No. is of the form 4K+3
=>(4K+3)² = 4e +3² = 4f +1 Remainder is 1
Hence, Proved.
----------------------------------------------------------------------------------------
Now it is given that
x² + y² = 2007 = 2004 + 3 = 4K +3 (where k = 2004/4)
We know that any perfect square when divided by 4 gives remainder either 0 or 1
So
x² + y² = 4K (if Both are x and y are even)
or x² + y² = 4K +1 (if one is even and other is odd)
or x² + y² = 4K +2 (if Both are odd)

As 2007 give remainder other than 0,1,2 so no integral values of (x,y) will be possible

Now your proof looks nice

 

[mastercool8695]

Re: mathematic related questions

The number of different positive integral solutions (x,y,z) such that x+y+z = 10

A) 36
B) 121
C) 10[SUP]3[/SUP] -10
D) [SUP]10[/SUP]C[SUB]3[/SUB] - [SUP]10[/SUP]C[SUB]2[/SUB]

offtopic : do you guys know what is the shortcut key for subscript and superscript here in this forum ?
have to go advanced everytime