254) If x^2-y^2=a where x,y are positive integers and a is an odd integer. If (x,y) has only one solution then a is
a) prime
b) composite
c) prime square
d) a ,b,c
e) a,c
255) If m^2+24m+21 has 3 factors for a natural number m, find the sum of all such m's
256) Find the number of pairs of positive numbers (x,y) such that x^3-y^3=100 and (x-y) and xy are integers.
a) 0 b) 1 c) 2 d) 4 e) none of these
Is the answer for 254 as
ReplyDeletea) prime number
b) the m has 2 values and they are 10 and 55
no correct option is e)
ReplyDeleteHi...
ReplyDeleteI could not solve the first problem.
255]
As it has 3 factors.. it is square of some prime number...
we get equation as (m+12+k) (m+12-k) = 125... so we get the factors 1,5,25,125... If we consider 25 (1*5*5)we get m = 3...
256]Answer is option 1... no pair will satisfy the value...
Can you help me with solutions?
@Rashmi here are the solutions
ReplyDelete255
for any natural number N=P^a.Q^b.R^c.. where P,Q,R are distinct prime numbers and
a,b,c are integers
no of factors is (a+1)(b+1)(c+1)
now as the no of factors is 3 one of a,b,c=2 and rest are zero
hence N is a prime square
let m^2+24m+21=p^2
then (m+12)^2+21-144=p^2
(m+12+p)(m+12-p)=123
m+12+p=123,41 m+12-p=1,3
2(m+12)=124,44
m=50,10
we check p=61,19 which are primes, hence both solution work
sum is 50+10=60
256)x^3-y^3=(x-y)[(x-y)^2+3xy]
let x-y=m and xy=n
then 100=m^3+3mn
clearly m<=4 and also m is a factor of 100
hence m=1,2,4
Case I when m=1
n=33
x-y=1 and xy=33
x^2 - x - 33 = 0
solving and taking positive value
x = (1+√(133))/2. Similarly we can find y.
case 2 m=2 n=46/3 not an integer
rejected
case 3 m=4 n=3
Since x - y = 4 we get x - 3/x = 4
Hence x^2 - 4x - 3 = 0
Solving we get x = (4 + √28)/2 = 2 + √7
hence we get two solutions
one for m=1 and other for m=4
hence two pairs
254) If x^2-y^2=a where x,y are positive integers and a is an odd integer. If (x,y) has only one solution then a is
ReplyDeletea) prime
b) composite
c) prime square
d) a ,b,c
e) a,c
(x-y)(x+y) = a
Since a is an odd integer both the terms of the product chould be odd. Which is possible when 1 of the x/y is even and other is odd.
Since(x,y) has only one solution => that a can be written as product of 2 nos in only 1 way which is possible only when a is a prime number.
Assuming a to be positive. x>y
x-y = 1
x+y=p
x=(p+1)/2
y=(p-1)/2
I am not sure about prime square. But if it is square of prime no then a can be written as 1,p^2 or p,p
Here p,p is not possible as in that case y = 0 which is not possible as it is positive.
Is it correct, outtimed?
@ Amit
ReplyDeletethe correct answer is e) that is both prime and prime square !
for prime you have proved
for prime square see this
(x+y)(x-y)=p^2
so =(p^2+1)/2 and (p^-1)/2
Great work, outtimed. Thank you very much for this blog.
ReplyDeleteCan we have more problems based on Concepts?
I think just trying to solve problems without having knowledge of concept doesn't help much. Concepts followed by problems is really great idea. I will be following the blog regularly from now on. :-)
@ Amit, yeah for sure I will be putting up more lessons, and every lesson will follow with more questions based on those lessons.
ReplyDeleteBut there are also people who are good in concepts just need enough material to challenge them, so for them we have problems of the week and puzzles !
Thanks!
Have Fun !
Hi outtimed ,
ReplyDeleteDude!!! Thanks for the explanation ...
Did some silly mistakes like 19 instead of 21 .. Phew...
Anyways...
Good Blog... Thanks to my friend who told me about it...
Keep good work!!!