Posts and Comments

ujjumba Saysis ans is 128??

I also got 128,

@Tallzz
17

Posted 28 Jul '10

> 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 SaysHi 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

@Tallzz
17

Posted 28 Jul '10

> 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

@Tallzz
17

Posted 28 Jul '10

> 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?

@Tallzz
17

Posted 28 Jul '10

> 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 SaysCan 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

- 1 Like

@Tallzz
17

Posted 28 Jul '10

> 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

@Tallzz
17

Posted 28 Jul '10

> 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 SaysIf 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

@Tallzz
17

Posted 20 Jul '10

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

ShivamSudarshan SaysIts c) both a and b

It should be c)

@AgentA

I got what u wanted to say

@Tallzz
17

Posted 20 Jul '10

> 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 SaysThere 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

@Tallzz
17

Posted 17 Jul '10

> *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!

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