@bodhi_vriksha said:How?
when n=0 => k=0
@Narci said:when n=0 => k=0
I mean the method buddy? 😃
The whole point of this problem is to be able to prove that no integer solution other than (0,0) exists.
@chillfactor said:
n(3n - 1) is a perfect sqWe know that n and (3n - 1) are co-prime to each other as 3(n) - (3n - 1) = 1So, we can say that for n(3n - 1) to be a perfect sq, both n and (3n - 1) has to be perfect squares, but (3n - 1) can never be a perfect square.Hence no such (k, n) are possible except (0, 0)
Nicely played chillfactor! That's the wholesome proof.
@bodhi_vriksha said:I mean the method buddy? The whole point of this problem in to be able to prove that no integer solution other than (0,0) exists.
n and 3n-1 are coprime to each other..
so both should be perfect squares..
and 3n-1 cannot be a perfect square as the square of an even number is always of the from 3n+1...
hence...
@Narci said:n and 3n-1 are coprime to each other.. so both should be perfect squares..and 3n-1 cannot be a perfect square as the square of an even number is always of the from 3n+1...hence...
First part of your explaination is fine but not the second part.
The reason for 3n-1 not being a perfect square is incorrect. For instance, 6 is an even number and its square is not of the form 3n+1.
All you need to understand is that any perfect square can be expressed as 3k or 3k+1 but not 3k-1. Why?
If b is an integer b can be of the form 3m, 3m+1, 3m-1
=> b^2 can be of the form 3k or 3k+1
@bodhi_vriksha said:First part of your explaination is fine but not the second part.The reason for 3n-1 not being a perfect square is incorrect. For instance, 6 is an even number and its square is not of the form 3n+1.All you need to understand is that any perfect square can be expressed as 3k or 3k+1 but not 3k-1. Why?If b is an integer b can be of the form 3m, 3m+1, 3m-1 => b^2 can be of the form 3k or 3k+1
yes, you are right... all even numbers except 6..
if we take b = 3m, then it will be of the from of 3k
if we take b = 3m+1 , then it will be of the from of 3k+ 1
if we take b = 3m-1, then also it will be of the form of 3k+1
What is the remainder when 3^1001 is divided by 1001? (try solving without using the CRT)
@Narci said:yes, you are right... all even numbers except 6..
Nope. All even numbers except those of the form 6k.
@bodhi_vriksha said:Nope. All even numbers except those of the form 6k.
ya.. that was explained in the last 3 lines of my previous post..
@bodhi_vriksha said:What is the remainder when 3^1001 is divided by 1001? (try solving without using the CRT)
3^6 when divided by 7 remainder is 1
=> 3^(6k) will leave remainder 1 by 7
3^10 when divided by 11 remainder is 1
=> 3^(10k) will leave remainder 1 by 11
3^6 when divided by 13 remainder is 1
=> 3^(6k) will leave remainder 1 by 13
=> 3^(6k) will leave remainder 1 by 13
So, we can say that 3^(30k) {30 is LCM of 6 and 10} will leave remainder 1 when divided by 1001
Hence, 3^1001 will leave remainder 3^11 or (2187*81) or (185*81) or 971
@ankitmishraiim said:iss wale ka koi hint dedo.. honi raha,,,
Couple of ways of doing this one...
I. Use Cauchy-schwarz €™s inequality, which says (a^2 + b^2)*(c^2 + d^2) >= (ac + bd)^2
((x-3)^2 + (y+2)^2)*(3^2 + 4^2) >= (3(x-3) + 4(y+2))^2
=> 9*25 >= (3(x-3) + 4(y+2))^2 => 3x + 4y
II. We have a circle with centre(3,-2) and radius 3. So for 3x+4y to be maximum, straight line 3x + 4y =c must be a tangent to this circle
Hence , |3*3 - 4*2 -c|/5 = 3 =>c(max) =16
A point P inside an equilateral triangle, ABC is located at a distance of 3,4 and 5 from A, B , C respectively. What is the area of the triangle.
@iLoveTorres try karo!! :)
@iLoveTorres try karo!! :)
A={179,180,181,...360}.B is a subset of A such that the sum of no two elements of B is divisible by 9.The number of elements in B cannot exceed:
a)81
b)82
c)85
d)101
e)102
a)81
b)82
c)85
d)101
e)102
@chillfactor said:3^6 when divided by 7 remainder is 1=> 3^(6k) will leave remainder 1 by 73^10 when divided by 11 remainder is 1=> 3^(10k) will leave remainder 1 by 113^6 when divided by 13 remainder is 1=> 3^(6k) will leave remainder 1 by 13So, we can say that 3^(30k) {30 is LCM of 6 and 10} will leave remainder 1 when divided by 1001Hence, 3^1001 will leave remainder 3^11 or (2187*81) or (185*81) or 971
Very good attempt!
Anyone with a different approach other than this one and the CRT?
@ChirpiBird said:A point P inside an equilateral triangle, ABC is located at a distance of 3,4 and 5 from A, B , C respectively. What is the area of the triangle.@iLoveTorres try karo!!
A property says that the sum of the distance of a point fom 3 sides in a equilateral triangle is equal to the altitude.
So altitude = 3+4+5 =12
So side = 8 ˆš3
Area = 48 ˆš3
So altitude = 3+4+5 =12
So side = 8 ˆš3
Area = 48 ˆš3
@saurabhlumarrai said:A={179,180,181,...
Consider values modulo 9, then take all 2,3,4,8 and leave out all 1,5,6,7 and finally take one mod 0. Total =82
@ChirpiBird said:A point P inside an equilateral triangle, ABC is located at a distance of 3,4 and 5 from A, B , C respectively. What is the area of the triangle.@iLoveTorres try karo!!
@chillfactor sir the answer 1.5 is correct......bt i'm unable to understand the soln. can u plz explain it
@catahead said:A property says that the sum of the distance of a point fom 3 sides in a equilateral triangle is equal to the altitude.So altitude = 3+4+5 =12So side = 8 ˆš3Area = 48 ˆš3
This is Viviani's theorem that you are talking about but you cannot apply it here since the problem provides distances from the vertices of the triangle and not the perpendicular distances from each side.
Try again please.