@chillfactor said:It can take all values greater than 0 and less than 360
Perfect 😃 
(except 180 of course)
regards
scrabbler
(except 180 of course)regards
scrabbler
@subhakimi said:Last one for the night ...Which is the smallest positive integer which can be represented as the sum of two perfect squares in exactly two ways ?
50
5^2+5^2 nd 7^2+1^2...dis can be if 2 perfect squares are nt distinct!!
5^2+5^2 nd 7^2+1^2...dis can be if 2 perfect squares are nt distinct!!
@mohitjain said:505^2+5^2 nd 7^2+1^2...dis can be if 2 perfect squares are nt distinct!!
in that case the number can be even less .... for 25 = 3^2 + 4^2 = 5^2 + 0^2
@subhakimi said:Bache ki na lo sir... positive number bolna bhul gaya toh itne easily kar loge aap Anyways I was looking for 65 actually.No one posted approach ... it can be done by hit & trial , I assume most had done that Try to give it a logical explanation as well @scrabbler , will think about this ... anyways good night for now
Any prime number of form 4k + 1 can be written as sum of two distinct positive perfect squares in exactly one way. Same is true for their squares also
So, 5, 13, 17, 25, 29, ... are such numbers
Similarly, product of any two numbers from the above list can be written as sum of two distinct positive perfect squares in exactly two ways.
Ex - 65, 85, 125, 145, 221, ....
Reason:-
(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2 = (ac - bd)^2 + (ad + bc)^2
@chillfactor said:Any prime number of form 4k + 1 can be written as sum of two distinct positive perfect squares in exactly one way. Same is true for their squares alsoSo, 5, 13, 17, 25, 29, ... are such numbersSimilarly, product of any two numbers from the above list can be written as sum of two distinct positive perfect squares in exactly two ways.Ex - 65, 85, 125, 145, 221, ....Reason:-(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2 = (ac - bd)^2 + (ad + bc)^2
Sir, Then what could be the probability that such numbers are divisible by 5 ?? ..
@ScareCrow28 said:Sir, Then what could be the probability that such numbers are divisible by 5 ?? ..
Since 5 is just one prime of infinite primes I would say that probability of the number being a multiple of 5 will be very less (almost insignificant)
@chillfactor said:Since 5 is just one prime of infinite primes I would say that probability of the number being a multiple of 5 will be very less (almost insignificant)
And the probability that N = a^2 + b^2 would be 9/25, right sir??
OR I am making a mistake somewhere! ??
But, we can't find the probability if N = a^2 + b^2 = c^2 + d^2 !!
@ScareCrow28 said:And the probability that N = a^2 + b^2 would be 9/25, right sir??OR I am making a mistake somewhere! ?? But, we can't find the probability if N = a^2 + b^2 = c^2 + d^2 !!
I didn't get you !!
Probability of what ??
Probability of what ??
@chillfactor said:I didn't get you !!Probability of what ??
Probability that a number can be expressed by sum of squares of 2 numbers, sir
@ScareCrow28 said:Probability that a number can be expressed by sum of squares of 2 numbers, sir
Nahi yaar, iski probability kaise nikali ??? yaar squares are not uniformly distributed, so I guess iski probability nahi nikaal sakte
@chillfactor said:Nahi yaar, iski probability kaise nikali ??? yaar squares are not uniformly distributed, so I guess iski probability nahi nikaal sakte
Sir jee:
a^2 mod 5 = 1, 0 or 4 only
So, a^2 + b^2 mod 5 can be obtained ..?? Am I wrong sir?
@ScareCrow28 said:Sir jee:a^2 mod 5 = 1, 0 or 4 onlySo, a^2 + b^2 mod 5 can be obtained ..?? Am I wrong sir?
Ah sorry my bad !! yeah you are right
Consider unit digits of a and b
i) both 0 or 5
ii) One 0 and other 5
iii) for one its 1 or 4 or 6 or 9 and for other its 2 or 3 or 7 or 8
So, probability = (2 + 2 + 4*4*2)/100 = 36/100 = 9/25
@chillfactor said:Ah sorry my bad !! yeah you are rightConsider unit digits of a and bi) both 0 or 5ii) One 0 and other 5iii) for one its 1 or 4 or 6 or 9 and for other its 2 or 3 or 7 or 8So, probability = (2 + 2 + 4*4*2)/100 = 36/100 = 9/25
\\___o/ Sir, Pranaam sweekar kare! Aur pareshan nahi karoonga 😃 Thanks for bearing with me, sir!
hello everybody. i'm varatharaj doing pre final year in mechanical engineering in coimbatore(tamilnadu).Right from the day i know about cat i started dreaming being an IIM er! MBA is my passion. im preparing for CAT 2013. As i was from a lower middle class background i cant afford to pay the coaching class fee. now iam preparing for CAT using the books from my college library.
but those books are really confusing and deviated from the actual CAT questions.i dont have a pc too. so i cant refer any online materials. i spend 4 hours daily for preparing CAT. my friend advised me that 4 hours of such preparations=1 hour of preparation by using standard materials like TIME,IMS. i went to TIME and ask them whether i can get the materials alone for a cheaper price. they said they will provide material only along with a course. and that course costs 20000! i was heart broken! i desperately need some standard books. i would be greatly thankful if someone is will to donate extra materials or old materials! PG is my only hope. so please help me! thank you all
but those books are really confusing and deviated from the actual CAT questions.i dont have a pc too. so i cant refer any online materials. i spend 4 hours daily for preparing CAT. my friend advised me that 4 hours of such preparations=1 hour of preparation by using standard materials like TIME,IMS. i went to TIME and ask them whether i can get the materials alone for a cheaper price. they said they will provide material only along with a course. and that course costs 20000! i was heart broken! i desperately need some standard books. i would be greatly thankful if someone is will to donate extra materials or old materials! PG is my only hope. so please help me! thank you all
How many right triangles with integral sides have one leg of length 35cm?
35^2=a^2-b^2=(a+b)(a-b)
a+b and a-b both have to be odd
odd factors of 35^2=
35*35=7^2*5^2=3*3
Remove 1 factor where there will be a+b=a-b=35
9-1=8
8/2 (remove cases where (a+b)
4
A cube has 8 vertices. For each pair of distinct vertices, we connect them up with a line segment. There are C(8 , 2) = 28 such line segments. For each of these 28 line segments, we mark the midpoint. How many distinct points have been marked as the midpoints ?
@jain4444 said:A cube has 8 vertices. For each pair of distinct vertices, we connect them up with a line segment. There are C(8 , 2) = 28 such line segments. For each of these 28 line segments, we mark the midpoint. How many distinct points have been marked as the midpoints ?
28-6-3=19??
@jain4444 said:A cube has 8 vertices. For each pair of distinct vertices, we connect them up with a line segment. There are C(8 , 2) = 28 such line segments. For each of these 28 line segments, we mark the midpoint. How many distinct points have been marked as the midpoints ?
19