[arya5139]

Originally Posted by MaskedMenace

1.10c3 = 120

2.

put 2^100 = (-1)

2^4 - 2^3 + 2^2 - 2 + 1

16 - 8 + 4 - 2 + 1

11

can u plz elaborate the line in bold?

[prateek7563]

What is the remainder obtained on dividing

16^101 + 8^101 + 4^101 + 2^101 + 1 by 2^100 +1

please solve this one. How can u think that substitution. please explain

[MaskedMenace]

Originally Posted by chandan.mnnit

Just tried in another way...tell me if its correct approach or not

total number of ways is 4!

in this each digit will occur at each place 4!/4 times that is 6 times

so at units place sum = 6* (1+2+3+4) =60

similarly we can find out for each of the tens, hundreds and thousands place.......Is this approach correct?

yes, that's correct
I guess there is a formula for this,

Sum = n!*(111..n ones)*(1+2+3+4+..n)

[effort]

Originally Posted by chandan.mnnit

Just tried in another way...tell me if its correct approach or not

total number of ways is 4!

in this each digit will occur at each place 4!/4 times that is 6 times

so at units place sum = 6* (1+2+3+4) =60

similarly we can find out for each of the tens, hundreds and thousands place.......Is this approach correct?

u have solved almost whole of the question,likewise it will same for ten's and hundred's place i.e 60.
so total is 60
60
60
--------------
6660
--------------

[MaskedMenace]

Originally Posted by prateek7563

What is the remainder obtained on dividing

16^101 + 8^101 + 4^101 + 2^101 + 1 by 2^100 +1

please solve this one. How can u think that substitution. please explain

It is the application of remainder theorem.

[randeepsingh]

hi puys .. this may sound naive to you but i have a doubt here ..

we say the area of triangle(attached) can be expressed as =

1/2 *bc sinA = 1/2 * ab sinC = 1/2 * ac sinB


can anyone please prove this for me ..

and also is Trignometry restricted to rt angles only ??



PS : I am not able to upload the pic of the triangle, dunno why its giving an error .. please consider ab,bc and ac as the sides of a triangle opposite to angles C,A and B respectively.

[arya5139]

Originally Posted by prateek7563

What is the remainder obtained on dividing

16^101 + 8^101 + 4^101 + 2^101 + 1 by 2^100 +1

please solve this one. How can u think that substitution. please explain

consider 2^101%(2^100 + 1) = (2^100 + 2^100)%(2^100 + 1)
= (2^100 +1 +2^100 +1 -2)%(2^1001)= -2

=>2^101 can be substituted by -2
=>given number =(2^101)^4 + (2^101)^3 + (2^101)^2 + 2^101 +1
=> remainder = -2^4 + -2^3 + -2^2 + -2^1 +1
=16-8+4-2+1=11

hope it clears..

[chandan.mnnit]

Originally Posted by randeepsingh

hi puys .. this may sound naive to you but i have a doubt here ..

we say the area of triangle(attached) can be expressed as =

1/2 *bc sinA = 1/2 * ab sinC = 1/2 * ac sinB


can anyone please prove this for me ..

and also is Trignometry restricted to rt angles only ??



PS : I am not able to upload the pic of the triangle, dunno why its giving an error .. please consider ab,bc and ac as the sides of a triangle opposite to angles C,A and B respectively.

This link might prove useful
Area of a triangle (7 different ways)

and no trignometry is not restricted to right angled triangles only

Hope i am not breaking any rules by posting the link here

[MaskedMenace]

Originally Posted by prateek7563

i didnt understand the solution.

_,_,_

12y => y can be {3,4,...9} 7 ways

13y =>6 ways
.
.
18y => 1 ways

total : 21ways

23y => y can be { 4,5,..9} => 6 ways
.
.
28y => 1 way

total : 15 ways

and so on. for 34y,45y,56y,67,78y.

[arya5139]

Originally Posted by randeepsingh

hi puys .. this may sound naive to you but i have a doubt here ..

we say the area of triangle(attached) can be expressed as =

1/2 *bc sinA = 1/2 * ab sinC = 1/2 * ac sinB


can anyone please prove this for me ..

and also is Trignometry restricted to rt angles only ??



PS : I am not able to upload the pic of the triangle, dunno why its giving an error .. please consider ab,bc and ac as the sides of a triangle opposite to angles C,A and B respectively.

consider any triangle ABC

drop perpendicular BD from point B to side AC

now, BD = AB*SINA

area of triangle = 1/2*BD*AC = 1/2*AB*SINA*AC=1/2*AB*AC*SINA=1/2*b*c*sinA

similarly others can also be proved

hope it clears...