@ScareCrow28 said:Find all integers which can be represented as a dierence of perfect squares, i.e.all m = x^2 ˆ' y^2.
Will it be infinite???
@AsihekAdhvaryu said:Will it be infinite???
Well, obviously there will be infinite possible combinations. But the question asked for a general answer, I suppose something's missing and deleted the Q.
@ScareCrow28 said:Sorry I think I missed something..
Yeah . I am not sure with the negative numbers 
A grasshopper lives on a coordinate line. It starts o at 1. It can jump either 1 unit or
5 units either to the right or to the left. However, the coordinate line has holes at all points
with coordinates divisible by 4 (e.g. there are holes at -4, 0, 4, 8 etc.), so the grasshopper
can not jump to any of those points. Can it reach point 3 after 2003 jumps?
5 units either to the right or to the left. However, the coordinate line has holes at all points
with coordinates divisible by 4 (e.g. there are holes at -4, 0, 4, 8 etc.), so the grasshopper
can not jump to any of those points. Can it reach point 3 after 2003 jumps?
P.S. Support your answer with reasoning
@ScareCrow28 said:A grasshopper lives on a coordinate line. It starts o at 1. It can jump either 1 unit or5 units either to the right or to the left. However, the coordinate line has holes at all pointswith coordinates divisible by 4 (e.g. there are holes at -4, 0, 4, 8 etc.), so the grasshoppercan not jump to any of those points. Can it reach point 3 after 2003 jumps?P.S. Support your answer with reasoning
let 5 unit step be x and 1 unit be y..
x+y=2003
5x-y=2
This won't give any integer solution...
@ScareCrow28 said:A grasshopper lives on a coordinate line. It starts o at 1. It can jump either 1 unit or5 units either to the right or to the left. However, the coordinate line has holes at all pointswith coordinates divisible by 4 (e.g. there are holes at -4, 0, 4, 8 etc.), so the grasshoppercan not jump to any of those points. Can it reach point 3 after 2003 jumps?P.S. Support your answer with reasoning
For first time to reach 3 it will need even steps.... but now its left with only odd steps so in cant be back to 3 again.......
@ScareCrow28 said:A grasshopper lives on a coordinate line. It starts o at 1. It can jump either 1 unit or5 units either to the right or to the left. However, the coordinate line has holes at all pointswith coordinates divisible by 4 (e.g. there are holes at -4, 0, 4, 8 etc.), so the grasshoppercan not jump to any of those points. Can it reach point 3 after 2003 jumps?P.S. Support your answer with reasoning
Every step it moves 1, 5, -1 or -5 i.e. an odd number. So in 2003 steps it will have moved the sum of 2003 odd numbers, which is again an odd number.
But to reach 3 it has to move an even number of steps.
So, not possible. Parity mismatch!
regards
scrabbler
But to reach 3 it has to move an even number of steps.
So, not possible. Parity mismatch!
regards
scrabbler
@ScareCrow28 never possible... since each jump can be of +1 ,-1, +5 , -5 distance so taking in terms of mod 4 we can have net displacemnt of 1(mod4) 0r -1(mod4) at each step.Now the current position is 1(mod4) thus it can move to 2(mod4) or 0(mod4) .each time u see the effective position in mod 4 will take value 0,1,2 or 3 & avoiding positions with 0(mod4) thus the only possible movement from 1(mod 4) to 3(mod4) will need two positive +1(mod 4) steps ..hence to reach at 3 (which is 3(mod4) ) will always take an even no of steps ..thus 2013 is not possible !!
Find the area of the region consisting of those points (x,y) for which x^2+y^2-4xx^2+y^2-8x1)2*(2pi-1)
2)4(pi/3 - _/3/2)
3) 4(2pi/3 - _/3/2)
4) 2(pi-2)
2)4(pi/3 - _/3/2)
3) 4(2pi/3 - _/3/2)
4) 2(pi-2)
@Buck.up said:Find the area of the region consisting of those points (x,y) for which x^2+y^2-4xx^2+y^2-8x1)2*(2pi-1)2)4(pi/3 - _/3/2)3) 4(2pi/3 - _/3)4) 2(pi-2)
3?
@Buck.up said:Find the area of the region consisting of those points (x,y) for which x^2+y^2-4xx^2+y^2-8x1)2*(2pi-1)2)4(pi/3 - _/3/2)3) 4(2pi/3 - _/3)4) 2(pi-2)
Option B..
@Buck.up said:its option 3falcao :can you throw more light.
these 2 are equations of two circles.u have to plot both of them on a coordinate plane.first circle has the centre at (2,0) and radius of 2 units whereas the other circle has a centre at (4,0) and has a radius of 2 units.the 2nd circle passes through the center of first circle i.e 2,0.the region that is being asked is the region formed by the intersection of these 2 circles.they meet at points (3,sqrt3) and (3,-sqrt3).so we have to calculate the area between points (2,0) (3,sqrt3) (4,0) and (3,-sqrt3).which comes out to be 3rd option
@Devanki said:The line AB is 6cm in length and is tangent to the inner of the concentric circles at point C. It is known that the radii of the two circles are integers. The radius of the outer is ------, where A and B are points on the outer circle5463
5 ?
@Buck.up said:The remainder of 11^11^11/9 ?P.S.Method other than Euler will be appreciated
5 imo. method same as scrabbler.
@Devanki said:The line AB is 6cm in length and is tangent to the inner of the concentric circles at point C. It is known that the radii of the two circles are integers. The radius of the outer is ------, where A and B are points on the outer circle5463
bigger radius= 5 small one=4.
@Buck.up said:Find the area of the region consisting of those points (x,y) for which x^2+y^2-4xx^2+y^2-8x1)2*(2pi-1)2)4(pi/3 - _/3/2)3) 4(2pi/3 - _/3)4) 2(pi-2)
none of these. ans is 4(2pi/3-_/3/2). kindly let me know.
@zuloo said:none of these. ans is 4(2pi/3-_/3/2). kindly let me know.
I guess you are doing a little mistake. OA is option 3. Check post#30177 for solution.
@Buck.up said:I guess you are doing a little mistake. OA is option 3. Check post#30177 for solution.
may be but according to me these are 2 circles with centres (2,0) & (4,0) and radius 2 which they are passing through each other's centre. accordingly,
2(pi*(2)^2* 120/360- 1/2*1*2_/3) so the answer comes out which i have written. just go through this and let me know again if there is any calculation mistake.
@zuloo said:may be but according to me these are 2 circles with centres (2,0) & (4,0) and radius 2 which they are passing through each other's centre. accordingly,2(pi*(2)^2* 120/360- 1/2*1*2_/3) so the answer comes out which i have written. just go through this and let me know again if there is any calculation mistake.
Yes. You are right. Ans comes out to be 4(2pi/3-_/3/2). I have edited the option in the ques.