For how many positive integer values of c does the equation 2x^2 + 689x + c = 0 have an integer solution ?
@jain4444 said:For how many positive integer values of c does the equation 2x^2 + 689x + c = 0 have an integer solution ?
344??
find the highest power of 567 that can divide ide 285! exactly..explain
@raopradeep said:the average weight of a group of n people is 75 kg.three men whose weights are 80,76,74 kg join the group and one man weighing between 90 to 100kg leave the group as a result average came down by 2 kg . if the number of men initially is a perfect square .then the weight of men who left the group is ?92949698
92
Find the largest integer S which is a divisor of n^5 ˆ' 17n^3 + 16n for every integer n ‰Ľ4.
@pavimai said:find the highest power of 567 that can divide ide 285! exactly..explain
35 i think
567 is 3^4*7
there are 35 number of 3^4 in 285!
and 45 numbers of 7
so i will go for 35
567 is 3^4*7
there are 35 number of 3^4 in 285!
and 45 numbers of 7
so i will go for 35
@pavimai said:find the highest power of 567 that can divide ide 285! exactly..explain
567=3^4*7
Highest power of 3 = 140 ---->Highest power of 3^4 = 35
Highest power of 7= 45
Therefore,highest power of 567 = min(35,45) = 35
Divide 3903 between amar and akbar such that amars share at the end of 7 yrs is equal to akbars share at the end of 9 yrs at 4 percent pa rate of compound interest
Ans is amar is 2028 and akbar is 1875
Ans is amar is 2028 and akbar is 1875
plz help...
@jain4444 said:Find the largest integer S which is a divisor of n^5 ˆ' 17n^3 + 16n for every integer n ‰Ľ4.
6??
n^5 - 17n^3 + 16n = n^5 - 16n^3 -n^3 + 16n = n^3 ( n^2 - 16) - n (n^2 -16 )
(n^3 -n) (n^2 - 16 ) = n (n+1)(n-1) (n+4)(n-4) = [(n-1)*n * (n+1) ] is always divisible by 6. ;
No..perfect rule for (n-4) (n+4 ) so it has to be 6?/
n^5 - 17n^3 + 16n = n^5 - 16n^3 -n^3 + 16n = n^3 ( n^2 - 16) - n (n^2 -16 )
(n^3 -n) (n^2 - 16 ) = n (n+1)(n-1) (n+4)(n-4) = [(n-1)*n * (n+1) ] is always divisible by 6. ;
No..perfect rule for (n-4) (n+4 ) so it has to be 6?/
@jain4444 said:For how many positive integer values of c does the equation 2x^2 + 689x + c = 0 have an integer solution ?
iska sol koi dal do ....
mera nahi a raha hai
mera nahi a raha hai
@jain4444 said:Find the largest integer S which is a divisor of n^5 ˆ' 17n^3 + 16n for every integer n ‰Ľ4.
24?
@joyjitpal said:iska sol koi dal do ....mera nahi a raha hai
integer solution ==> b^2 - 4ac has to be a perfect square
(689)^2 - 4*2*c = k^2 ==> (689+k)(689-k ) = 8*c ==> c = (689 + k) (689 -k) /8
we get C as a +ve integr for all odd values of k from 1 to 687..so total 344 values...correct me if i am wrong
@jain4444 sir a dekho..wats d OA..is dis the right approach??
(689)^2 - 4*2*c = k^2 ==> (689+k)(689-k ) = 8*c ==> c = (689 + k) (689 -k) /8
we get C as a +ve integr for all odd values of k from 1 to 687..so total 344 values...correct me if i am wrong
@jain4444 sir a dekho..wats d OA..is dis the right approach??
@ani6 said:Divide 3903 between amar and akbar such that amars share at the end of 7 yrs is equal to akbars share at the end of 9 yrs at 4 percent pa rate of compound interestAns is amar is 2028 and akbar is 1875
Write the equation and get the common terms out,
it then comes out easily
it then comes out easily
@raopradeep said:the average weight of a group of n people is 75 kg.three men whose weights are 80,76,74 kg join the group and one man weighing between 90 to 100kg leave the group as a result average came down by 2 kg . if the number of men initially is a perfect square .then the weight of men who left the group is ?92949698
sum of weights= 75*n----1
let x be the weight of the man who left the grp
sum of weights+80+76+74-x=73(n+2)
sum of weights+230-x=73(n+2)---2
subtract eq 1 from 2
73(n+2)-75n = 230 -x
x=230+75n-73(n+2)
n is the perfect square and x has to lie in the range 90 to 100..
possible value n=4
x=92
@ravi.theja said:integer solution ==> b^2 - 4ac has to be a perfect square(689)^2 - 4*2*c = k^2 ==> (689+k)(689-k ) = 8*c ==> c = (689 + k) (689 -k) /8we get C as a +ve integr for all odd values of k from 1 to 687..so total 344 values...correct me if i am wrong @jain4444 sir a dekho..wats d OA..is dis the right approach??
if the discriminant is a perfect square
we get rational roots but here the roots are integer :)
we get rational roots but here the roots are integer :)
@techsurge said:whats the OA, know only how to calculate less than 150
Numbers less than 150 co-prime with it = 150(1-1/2)(1-1/3)(1-1/5) = 40
Sum of such numbers = 150/2 * 40 = 3000
If a number x is co-prime with 150 ,then 150 + x too will be co-prime with it.
Therefore,sum = 3000 + 40*150 + 3000 =12000