@joyjitpal said:20^2004 + 16^2004 - 3^2004 -1 is divisible by a)317 b)323 c)253 d)91
323??
323 = 17*19
=> (3^2004 + (-1)^2004 - 3^2004 - 1)%17 = 0
and (1^2004 + (-3)^2004 - 3^2004 - 1)%19 = 0
hence, divisible by 323...
@saurav205 said:323
@Logrhythm said:323??323 = 17*19=> (3^2004 + (-1)^2004 - 3^2004 - 1)%17 = 0and (1^2004 + (-3)^2004 - 3^2004 - 1)%19 = 0hence, divisible by 323...
@joyjitpal said:20^2004 + 16^2004 - 3^2004 -1 is divisible by a)317 b)323 c)253 d)91
323..as (20^2004-1^2004) is always divisible by 19 and (16^2004-3^2004) is divisible by 19 as 2004 is even..so d whole no. should be divisible by 19..hence 323..=19*7
@joyjitpal said:20^2004 + 16^2004 - 3^2004 -1 is divisible by a)317 b)323 c)253 d)91
323 ?
323 = 17*19
20^2004 + 16^2004 - 3^2004 - 1 mod 17 = 0
20^2004 + 16^2004 - 3^2004 - 1 mod 19 = 0
Chalo mein hi question daal deta hun...
If x^4*y^2 = 400 and x and y are positive, then the least value of 5x + 8y is
A)40
B)30
c)25
D)None of these
A)40
B)30
c)25
D)None of these
@Logrhythm said:Chalo mein hi question daal deta hun...If x^4*y^2 = 400 and x and y are positive, then the least value of 5x + 8y isA)40 B)30 c)25 D)None of these
30 hoga
5X/4 = 8Y/2 = K
X = 4K/5
and Y=K/4
putting it in x^4*y^2 = 400
we get X=4 and Y=5/4
so 5x + 8y is 30
5X/4 = 8Y/2 = K
X = 4K/5
and Y=K/4
putting it in x^4*y^2 = 400
we get X=4 and Y=5/4
so 5x + 8y is 30
@Logrhythm said:Chalo mein hi question daal deta hun...If x^4*y^2 = 400 and x and y are positive, then the least value of 5x + 8y isA)40 B)30 c)25 D)None of these
30 ?
5x/4 = 8y/2
y = 5x/16
x^4 * y^2 = 2^4*5^2
x^4 * (5x/16)^2 = 2^4 * 5^2
x^6 = 2^12----> x = 4
y = 5*4/16 = 5/4
5(4) + 8(5/4) = 30
@Logrhythm said:Chalo mein hi question daal deta hun...If x^4*y^2 = 400 and x and y are positive, then the least value of 5x + 8y isA)40 B)30 c)25 D)None of these
30 aayega....
(5x/4 + ...4 terms + 8y/2 +8y/2 ) >= (6*(5/4)^4*(8/2)^2*400)^1/6
on solving 6*5 = 30
@Logrhythm said:Chalo mein hi question daal deta hun...If x^4*y^2 = 400 and x and y are positive, then the least value of 5x + 8y isA)40 B)30 c)25 D)None of these
5x/4=8y/2=K
Now it is easy
(b) 30
Now it is easy
(b) 30
@saurav205 said:30 aayega....(5x/4 + ...4 terms + 8y/2 +8y/2 ) >= (6*(5/4)^4*(8/2)^2*400)^1/6on solving 6*5 = 30
is it a.m>=g.m..can u plz xplain how 2 use dis property in this case?
@mohitjain said:is it a.m>=g.m..can u plz xplain how 2 use dis property in this case?
yes am >= gm
(5x/4 + 5x/4 + 5x/4 + 5x/4 + 8y/2 + 8y/2)/6 >= ((5x/4)^4*(8y/2)^2)^(1/6) = (5^6*2^10/2^10)^(1/6) = 6*5 = 30
@mohitjain said:is it a.m>=g.m..can u plz xplain how 2 use dis property in this case?
see , x as 5 as its coefficient and the power of x was 4..so divided 5x into four parts to get x^4 on the GM side.
same with y,
8 as the coefficient and power of 2..so divided 8y into 2 parts to get the power of y as 2
since 6 numbers are involved here 4 of x and 2 of y..so divided by 6 on the AM side....
Keep in mind the known data that is given to you.....that helps in deciding how you will break the parts. In this case it was x^4 and y^2..
hope this helps..
@saurav205 said:see , x as 5 as its coefficient and the power of x was 4..so divided 5x into four parts to get x^4 on the GM side.same with y,8 as the coefficient and power of 2..so divided 8y into 2 parts to get the power of y as 2since 6 numbers are involved here 4 of x and 2 of y..so divided by 6 on the AM side....Keep in mind the known data that is given to you.....that helps in deciding how you will break the parts. In this case it was x^4 and y^2..hope this helps..
thanx buddy..got it!!!!
@saurav205 @mailtoankit @joyjitpal
The quantities of a sugar solution in three different vessels form an arithmetic progression with a common difference of 5 kg. All the three sugar solutions have the same concentration. The difference between the numerical values of the concentration of sugar in the sugar solution and the quantity of sugar solution in the vessel having the least quantity is 8. The total quantity of sugar in the three vessels is 5.4 kg. What are the quantities of sugar solution in the three vessels?
a 5 kg, 10 kg and 15 kg
b 6 kg, 11 kg and 16 kg
c 7 kg, 11 kg and 18 kg
d 8 kg, 13 kg and 18 kg
e 7 kg, 12 kg and 17 kg
The quantities of a sugar solution in three different vessels form an arithmetic progression with a common difference of 5 kg. All the three sugar solutions have the same concentration. The difference between the numerical values of the concentration of sugar in the sugar solution and the quantity of sugar solution in the vessel having the least quantity is 8. The total quantity of sugar in the three vessels is 5.4 kg. What are the quantities of sugar solution in the three vessels?
a 5 kg, 10 kg and 15 kg
b 6 kg, 11 kg and 16 kg
c 7 kg, 11 kg and 18 kg
d 8 kg, 13 kg and 18 kg
e 7 kg, 12 kg and 17 kg
@mohitjain
Shl be Option E: 7, 12 n 17..
Let the Amt of Vessels be x-5 kg, x kg n x+5 kg receptively..n let c% be the conc in each Vessel..
Given, c/100*(x-5 + x + x + 5)= 5.4, or, cx = 180...(i)
Again, c -(x-5) = 8, or, c-x=3, or, c = x+3...(ii)
Thus, x^2 + 3x - 180 = 0
=>(x+15)(x-12) = 0
Thus, x will be 12 kg..Thus, V1= 7 kg, V2= 12 kg n V3= 17 kg..