official quant thread for cat08

[schander2020]

Originally Posted by implex

You can post your answers, by posting your keys using the comment button !!
Best of Luck
Cheers!

Hi Implex,

I need a clarification on solution provided for Q.196 in Quant Prblm Set 3.
This is the question....
196) Let p(x) = x2 + 40. Then for any two positive integers i and j where
i > j, is p(i) + p(j) a composite number?
X: p(i)p(j) is not a composite number
Y: p(2i) + p(2j) is a composite number

And the solution starts like this....
p(i)p(j) is not a composite number
=> i2 −j2 is a prime as i, j are positive integers and i > j, (i2 −j2) can’t be
1.......

How the conclusion, "i2-j2 should be prime" was derived from the condition "p(i)p(j) is not a composite number"?

please answer.

[prade]

how to find the remainder when (2^133) is divided by 133

133 = 7 * 19
EN(133) = LCM(18,6) = 18

2^133 mod 133 = 2^7 mod 133 = 128 mod 133

[prvineeth]

Originally Posted by shivam_01

In this case take the
2^133 mod 7 and 2^133 mod 19 sepeartely

and the first one gives the remainder as 2 and the second one as 14

so we have
7x+2=19y+14
which gives the answer as 128

Am i correct ??

Which funda did you use here?Plz explain....

[prvineeth]

I have a serious doubt
Let N1=an+1,N2=bn+1,N3=cn+1........Is the remainder 1 when N1N2N3 is divided by n??

[proletarian]

double post...

[proletarian]

Originally Posted by prvineeth

I have a serious doubt
Let N1=an+1,N2=bn+1,N3=cn+1........Is the remainder 1 when N1N2N3 is divided by n??

N1N2N3=(an+1)(bn+1)(cn+1)=( acn^3 +an^2 +acn^2 +an+bcn^2+bn+cn+1)
so remainder=1

In general, the remainder of the product would be the product of its remainders

[prvineeth]

what is the remainder when 2^133 is divided by 133?
My solution......
133=7*19
2&133 coprimes........euler number(133)=6*18=108

Remainder(2^108/133)=1=> 2^108=a133+1 => 2^36=b133+1--------(i) since (2^36)(2^36)(2^36)=b133+1...............
2^18=c133+1 -------(ii)from eqn(i),
2^6=d133+1----------(iii) from eqn(ii)
=> (i)(ii)(iii) => 2^132=k133+1=> 2^133= x133+2
So remainder 2,......................how 128??

[proletarian]

Originally Posted by prvineeth

what is the remainder when 2^133 is divided by 133?
My solution......
133=7*19
2&133 coprimes........euler number(133)=6*18=108

Remainder(2^108/133)=1=> 2^108=a133+1 => 2^36=b133+1--------(i) since (2^36)(2^36)(2^36)=b133+1...............
2^18=c133+1 -------(ii)from eqn(i),
2^6=d133+1----------(iii) from eqn(ii)
=> (i)(ii)(iii) => 2^132=k133+1=> 2^133= x133+2
So remainder 2,......................how 128??

dude, the remainder of the products can be the product of the remainders but not the other way around. wrong assumption!
2^6 =64 can't be written as c133+1

[sidcool007]

How many integer soultions exist of the eqn 243x + 198y = 9 . (given 0<x<100)
1.1
2.3
3.5
4.6
5.7

Answer with the solution will be appreciated.Thanks.

[prakharc]

Originally Posted by prade

how to find the remainder when (2^133) is divided by 133

133 = 7 * 19
EN(133) = LCM(18,6) = 18

2^133 mod 133 = 2^7 mod 133 = 128 mod 133

Is this way correct.....can we take EN(133) like this ??