a) a, b, c are distinct.
b) a, b, c belongs to {1, 2, 3, …2008}
c) x + 1 divides ax² + bx + c
b) a, b, c belongs to {1, 2, 3, …2008}
c) x + 1 divides ax² + bx + c
Find the number of quadratic polynomials ax² + bx + c such that:
@bullseyes said:Find the number of quadratic polynomials ax² + bx + c such that:a) a, b, c are distinct.b) a, b, c belongs to {1, 2, 3, …2008}c) x + 1 divides ax² + bx + c
b=a+c
a+c solve 😁
@bullseyes said:Find the number of quadratic polynomials ax² + bx + c such that:a) a, b, c are distinct.b) a, b, c belongs to {1, 2, 3, …2008}c) x + 1 divides ax² + bx + c
Edit: something like 2014024?? 
@soumitrabengeri said:A positive integer is equal to the square of the number of factors it has. How many such integers are there?(a) 1(b) 2 (c) 3 (d) InfinitePlease explain with approach
Crux of the problem
consider two terms x^n and (n+1)^2 {where x is a prime number and n is +ve interger}
Now
if x=2 then x^n > (n+1)^2 if n>=6
if x=3 then x^n > (n+1)^2 if n>=3
if x>=5 then x^n > (n+1)^2 for all values of n
Solution
say N = (p1^a)*(p2^b)*(p3^c)..... where pi is prime factor.
no. of factors = (1+a)(1+b)(1+c)...
no. of factors = (1+a)(1+b)(1+c)...
As per given question we are looking for
(p1^a)*(p2^b)*(p3^c)..... = {(1+a)(1+b)(1+c)....}^2
(p1^a)*(p2^b)*(p3^c)..... = {(1+a)(1+b)(1+c)....}^2
based on the conclusions drawn above we can say that p1,p2,p3.... cannot be all greater than 3 at the same time. because if p1,p2,p3... are all greater than 3
then we get
p1^a > (1+a)^2
p2^b > (1+b)^2
p3^c > (1+c)^2
.
.
multiplying all we get (p1^a)*(p2^b)*(p3^c)..... > {(1+a)(1+b)(1+c)....}^2
hence N = 2^a*3^b*5^c.....
Also N is perfect square => all the powers must be even i.e a,b,c all must be even
=> possible combinations values for (a,b,c....) under the constraints
(0,0,0....), (0,2,0,0..)
hence only 2 possible values.
ATDH.
@bullseyes said:Find the number of quadratic polynomials ax² + bx + c such that:a) a, b, c are distinct.b) a, b, c belongs to {1, 2, 3, …2008}c) x + 1 divides ax² + bx + c
2008*1003 ... ordered solutions
as -1 is root of eq.. put -1 in eq.. we get
a+c=b
now
if a= 1 ... c can take any odd value from 1 to 2008 except 1.. 1003 values
if a=2.. c can take any even value from 1 to 2008 except 2.. 1003 values
.... so on till a=2008
@FoolNFinal said:2008*1003 ... ordered solutions as -1 is root of eq.. put -1 in eq.. we geta+c=bnow if a= 1 ... c can take any odd value from 1 to 2008 except 1.. 1003 valuesif a=2.. c can take any even value from 1 to 2008 except 2.. 1003 values.... so on till a=2008
or we can use:
a+c + x n - 1004
@FoolNFinal said:2008*1003 ... ordered solutions as -1 is root of eq.. put -1 in eq.. we geta+c=bnow if a= 1 ... c can take any odd value from 1 to 2008 except 1.. 1003 valuesif a=2.. c can take any even value from 1 to 2008 except 2.. 1003 values.... so on till a=2008
i think the answer is 2014024
@Brooklyn that's how u solve a question 😛
(no offence)
@bullseyes said:i think the answer is 2014024 @Brooklyn that's how u solve a question (no offence)
thats also one way
the way i meant was:
a+c + x= 2006
2008c2 - 1004 😃 solve it 😃 u'll get same ans 😃
@ScareCrow28 said:3
@Brooklyn said:thats also one waythe way i meant was:a+c + x= 20062008c2 - 1004 solve it u'll get same ans n none taken
@bullseyes said:i think the answer is 2014024 @Brooklyn that's how u solve a question
bhai .. 2008*1003 = 2014024
@Brooklyn ur ans is 1004*2007 .... which has cases where a=c also...
so we need to cancle that too...
@FoolNFinal said:bhai .. 2008*1003 = 2014024@Brooklynur ans is 1004*2007 .... which has cases where a=c also...so we need to cancle that too...
i never said its a wrong answer
@FoolNFinal said:bhai .. 2008*1003 = 2014024@Brooklynur ans is 1004*2007 .... which has cases where a=c also...so we need to cancle that too...
what i did was:
a+c
now a,c> 1
a+c
add dummy x
a+c + x=2006
total integeral soln: 2008 C 2
now
a=c is possible for even nos such as 2,4 so on
total even nos = 2008/2 = 1004 :)
@maneeshp said:when a is odd why c can take only odd value?
2 odd nos add to even no. not even + odd
check :
2+3 = 5 odd
2+2 =even
3+5 = even 😃
300! is divisible by (24!)^n.
Max. possible integral value of n ?
@maneeshp said:that i know...what i meant is if a is odd,why c also odd i.e is there any condition that b should be even?
sorry ask the persona then 😃 i cant understand his logic completely so cant ans
@bullseyes said:300! is divisible by (24!)^n.Max. possible integral value of n ?
13....
accidently mere ans sahi ho gaya..
w8 dekhata hu..