@vijay_chandola said:11/7, 11/7Take the mirror image of any one point w.r.t. y=x and then compare the slopes
Or check with the options :)
@kleptomaniac_20 said:@chillfactor : find the remainder when 50^51^52 is divided by 11?
E(11)=10
51^52 mod 10
=1 mod 10
=> 50^1 mod 11
= 6 mod 11
@kleptomaniac_20 said:@chillfactor : find the remainder when 50^51^52 is divided by 11?
E(11)=10
51^52 mod 10 = 1
50 mod 11 = 6
@kleptomaniac_20 said:@chillfactor : find the remainder when 50^51^52 is divided by 11?
Euler of 11 is 10
51^52 mod 10=1
50^51^52 mod 11=6
51^52 mod 10=1
50^51^52 mod 11=6
@sauravd2001 said:@deedeedudu hw cm check with the options yeh bta bhai
eg. Lets assume the point to b 10/9,10/9
AP= root[(1-10/9)^2+(-3-10/9)^2]
BP=root[(5-10/9^2+(2-10/9)^2]
Add both
Likewise do for all the options to see which one gives least value
AP= root[(1-10/9)^2+(-3-10/9)^2]
BP=root[(5-10/9^2+(2-10/9)^2]
Add both
Likewise do for all the options to see which one gives least value
@sauravd2001 said:Consider points A(1, €“3) and B(5, 2). Let P is a point on the line y = x. Find co-ordinates of P for which |AP + PB| is minimum.OPTIONS1) 10/9,10/9 2) 11/9,11/9 3) 11/7,11/7 4) 13/7,13/7
As, both points are on the same side of the line y = x
From any point on the y = x, point B and B' (mirror image) are equidistant.
Now, in the question we have to minimize AP + PB, as PB = PB', we can say that
We have to minimize AP + PB' (or AB'), so follow the straight line path and find the least possible distance.
Required point will be intersection point of AB' and y = x
Another question on similar logic (check the attachment)
@kleptomaniac_20 said:@chillfactor : find the remainder when 50^51^52 is divided by 11?
Euler's number of 11 is 10
and 51^52 when divided by 10 remainder is 1
So, remainder is 50^1 or 6
@chillfactor said:From any point on the y = x, point B and B' (mirror image) are equidistant.Now, in the question we have to minimize AP + PB, as PB = PB', we can say thatWe have to minimize AP + PB' (or AB'), so follow the straight line path and find the least possible distance.Required point will be intersection point of AB' and y = xAnother question on similar logic (check the attachment)Euler's number of 11 is 10and 51^52 when divided by 10 remainder is 1So, remainder is 50^1 or 6
28root2?
@chillfactor said:f(x) = p(x - a)(x - 3)g(x) = q(x - a)(x - 5)So, f(5) = 2p(5 - a)g(7) = 2q(7 - a)pq(5 - a)(7 - a) = 3So, none of these, as value of a depends on p and q
either pq=1 and (5-a)*(7-a)=3 or (5-a)(7-a)=1 or pq=3
First one gives solution.
a=4 or 8
@gnehagarg said:either pq=1 and (5-a)*(7-a)=3 or (5-a)(7-a)=1 or pq=3First one gives solution.a=4 or 8
It is nowhere mentioned that coefficients of the quadratic equations are integers. So, p and q can take any value
If the sum of ten non-negative integers is 48, what is the minimum value of the sum of the cubes of the ten integers?
@bullseyes Agar 0....(9times) aur 48 m divide kare to cube ki value min hogi i.e answer should be 110592.
@AIM_IIM_2013 said:@bullseyes Agar 0....(9times) aur 48 m divide kare to cube ki value min hogi i.e answer should be 110592.
@sauravd2001 said:@bullseyes 2052
no
@bullseyes said:If the sum of ten non-negative integers is 48, what is the minimum value of the sum of the cubes of the ten integers?
48/10=4.8
so, 9 numbers will be 5 and rest one will be 3
=> sum of cubes=5^3*9+3^3=1152
Kuchh gadbad h :neutral:
Let me check again :O
@bullseyes said:If the sum of ten non-negative integers is 48, what is the minimum value of the sum of the cubes of the ten integers?
Maximize the repeating digit such that remaining digits to add be min.
9(5)+3 fits
min sum=1152
In any other case it is comming more than that