Tallzz

@Tallzz

> ujjumba Says > > is ans is 128?? I also got 128,
ujjumba Says
is ans is 128??

I also got 128,
> saurav_chetry Says > > Hi Puys please help me out in finding the remainder of 2^133/133. Please inform me if there is any approach pertaining to the problems where the divisor and the power are same? options kya hain dude
saurav_chetry Says
Hi Puys please help me out in finding the remainder of 2^133/133. Please inform me if there is any approach pertaining to the problems where the divisor and the power are same?

options kya hain dude
> there is a shortcut but works only if divisor is prime number. Fermet's theorem: If p is prime number then n^p%p would always be n. So here answer would be 2. 133 = 7*19
there is a shortcut but works only if divisor is prime number.

Fermet's theorem: If p is prime number then n^p%p would always be n.

So here answer would be 2.

133 = 7*19
> its 122976 sum of AP ,last term 997 ,1st term 101, D = 4 :: Suja :: I got all this and total no of terms?
its 122976

sum of AP ,last term 997 ,1st term 101, D = 4

Suja

I got all this

and total no of terms?
> ABCLIKS Says > > Can u plz give the approach...? (a2+b2+c2)^2 = (a4+b4+c4+2(a2b2+b2c2+a2c2)) so, a4+b4+c4 = 25 - 2(a2b2+b2c2+a2c2) now, ab+bc+ca = 2(find out), so, (ab+bc+c)^2 = (a2b2+b2c2+a2c2+2(abc(a+b+c))=(a2b2+b2c2+a2c2 + 6abc) so, (a2b2+b2c2+a2c2) = 4-6abc we hav to f...
ABCLIKS Says
Can u plz give the approach...?

(a2+b2+c2)^2 = (a4+b4+c4+2(a2b2+b2c2+a2c2))
so, a4+b4+c4 = 25 - 2(a2b2+b2c2+a2c2)

now, ab+bc+ca = 2(find out),
so, (ab+bc+c)^2 = (a2b2+b2c2+a2c2+2(abc(a+b+c))=(a2b2+b2c2+a2c2 + 6abc)
so, (a2b2+b2c2+a2c2) = 4-6abc

we hav to find 6abc,
(a2+b2+c2)(a+b+c)=a3+b3+c3+(ab2+a2b+b2c+bc2+a2c+ac2)
so, (ab2+a2b+b2c+bc2+a2c+ac2)=8,
so,6abc=-4

so the answer is a4+b4+c4 = 25 - 2(a2b2+b2c2+a2c2)=25-16=9
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> What is the sum of all the three digit numbers which leave the remainder 1 when divided by 4? (a)123525 (b)122976 (c)112996 (d)100000 answer is a) 123525
What is the sum of all the three digit numbers which leave the
remainder 1 when divided by 4?
(a)123525 (b)122976 (c)112996 (d)100000

answer is a) 123525
> victory2010 Says > > If a+b+c=3 and a(^2)+b(^2)+c(^2)=5 and a(^3)+b(^3)+c(^3)=7 what is a(^4)+b(^4)+c(^4) ? is the answer 9
victory2010 Says
If a+b+c=3 and a(^2)+b(^2)+c(^2)=5 and a(^3)+b(^3)+c(^3)=7 what is a(^4)+b(^4)+c(^4) ?

is the answer 9
> ShivamSudarshan Says > > Its c) both a and b It should be c) @AgentA :: :: :: :: :: I got what u wanted to say
ShivamSudarshan Says
Its c) both a and b

It should be c)

@AgentA
I got what u wanted to say
> topper@IITK Says > > *There are 30 people and 3 clubs M, S, and Z in a company. 10 people joined M, 12 people joined S and 5 people joined Z. If the members of M did not join any other club, at most, how many people of the company did not join any club?*** I am getting 8 as the answer
topper@IITK Says
There are 30 people and 3 clubs M, S, and Z in a company. 10 people joined M, 12 people joined S and 5 people joined Z. If the members of M did not join any other club, at most, how many people of the company did not join any club?

I am getting 8 as the answer
> *4-10000! = (100!)K P, where P and K are integers. What can be the maximum value of K?* yaar K*P=10000x9999x9998x.......102x101.... obviously K max is 10000x9999x9998x.......102...!!!! is it 4-10000!
4-10000! = (100!)K P, where P and K are integers. What can be the
maximum value of K?


yaar K*P=10000x9999x9998x.......102x101....

obviously K max is 10000x9999x9998x.......102...!!!!

is it 4-10000!