@adityaknsit said:(3^1024)-1 =(2+1)^1024-1=1+2*(1024C1)+2^2*(1024C2)...............- 1.........now all will contain at least 1024*2= 2^11...........
y u not taking 4-1 = 3..by that em getting 12..can u justify between bth the approaches
If a and b are two integers, then find total no of solutions for this eqn.?
a!b! = a! + b!
PS: m not looking for answer but approach
On a river B is between A and C and equidistant from them.A boat goes from A to B and back in 5hr 15mins and from A to C in 7 hours.How long will it take to go from C to A ,if water flows from A to C.
options: 3hrs, 3.5hrs,4 hrs, 4.5 hrs
Plz tell approach.
@bullseyes said:If a and b are two integers, then find total no of solutions for this eqn.?a!b! = a! + b!
1 - 1/b! = 1/a!
1/A! + 1/B! = 1
by this only one solution (a,b) = (2,2)
1/A! + 1/B! = 1
by this only one solution (a,b) = (2,2)
On a river B is between A and C and equidistant from them.A boat goes from A to B and back in 5hr 15mins and from A to C in 7 hours.How long will it take to go from C to A ,if water flows from A to C.
options: 3hrs, 3.5hrs,4 hrs, 4.5 hrs
Plz tell approach.
@rkshtsurana said:y u not taking 4-1 = 3..by that em getting 12..can u justify between bth the approaches
That is what id on't understand....I took 2+1...i don't understand where does my approach leak?
@adityaknsit said:That is what id on't understand....I took 2+1...i don't understand where does my approach leak?
its from CL mock ... to find highest power.. take highest power of 2 in play..yes u can write 2+1 but if you take 4 -1...u taking4 in the scene which has higher power than 2 and ll definately give higher power of 2 overall... this was a tricky question and many of us did wrng initailly
@bullseyes
a!b! = a! + b!
now, b!= 1 + b!/a!
now, we know that all factorials above 1! are even........so.....either b!/a! is 0 or 1....no factorial can be 0.........only one alternative remains, that 1 + b!/a!=2
Now, for this.....B!=2 and a!=b!.......
hence a=b=2
a!b! = a! + b!
now, b!= 1 + b!/a!
now, we know that all factorials above 1! are even........so.....either b!/a! is 0 or 1....no factorial can be 0.........only one alternative remains, that 1 + b!/a!=2
Now, for this.....B!=2 and a!=b!.......
hence a=b=2
@bullseyes said:If a and b are two integers, then find total no of solutions for this eqn.?a!b! = a! + b! PS: m not looking for answer but approach
(2,2)...1 solution?
Did by putting values..
If sin(x-2y)=cos(4y-x),find cot 2y.
I am having problem in solving it.Can anyone give me some idea?
@anitito123 said:On a river B is between A and C and equidistant from them.A boat goes from A to B and back in 5hr 15mins and from A to C in 7 hours.How long will it take to go from C to A ,if water flows from A to C.options: 3hrs, 3.5hrs,4 hrs, 4.5 hrsPlz tell approach.
d/ (u + v) + d /(v -u) = 21/4
2d/ (u+v) = 7
from this u + v = 2d/7
substituting
d/(v-u) = 21/4 - 14/4 = 7/4
v-u = 4d/7
2v = 6d/7-------------> v =3d/7
u = -d/7
c->a
2d/v-u = 2d/(4d/7) = 14/4 = 7/2 = 3.5 hrs
2d/ (u+v) = 7
from this u + v = 2d/7
substituting
d/(v-u) = 21/4 - 14/4 = 7/4
v-u = 4d/7
2v = 6d/7-------------> v =3d/7
u = -d/7
c->a
2d/v-u = 2d/(4d/7) = 14/4 = 7/2 = 3.5 hrs
@Rakaesh said:If sin(x-2y)=cos(4y-x),find cot 2y.I am having problem in solving it.Can anyone give me some idea?
sin(x - 2y ) = cos( 4y - x)
cos( pie/2 - x + 2y) = cos( 4y - x)
pie/2 - x + 2y = 4y -x
pie/2 = 2y
cot2y = cot(pie/2) = 0
cos( pie/2 - x + 2y) = cos( 4y - x)
pie/2 - x + 2y = 4y -x
pie/2 = 2y
cot2y = cot(pie/2) = 0
@anitito123 said:On a river B is between A and C and equidistant from them.A boat goes from A to B and back in 5hr 15mins and from A to C in 7 hours.How long will it take to go from C to A ,if water flows from A to C.options: 3hrs, 3.5hrs,4 hrs, 4.5 hrsPlz tell approach.
3.5 hrs..
x/(vb+vc) +x/(vb-vc) =5.25
2x/(vb+vc)=7
so 2x/(vb-vc) = 3.5
x/(vb+vc) +x/(vb-vc) =5.25
2x/(vb+vc)=7
so 2x/(vb-vc) = 3.5
@krum said:7c2*7c1*(6*5*4*3*2)/7^7=>3*6!/7^5
Hey can you please explain how that factor of 6! is coming?