[naga25french]

Originally Posted by ganesh99

[/b] can someone pls explain this

133 = 19 *7

euler number = lcm of {(7-1),(19-1)} = Lcm of 6 and 18 = 18

2^18 leaves remainder 1 when divided by 133


so 2^18k leaves remainder 1


126 = 18 *7

so we r left with 2^7 divide by 133

so the remainder is 128

[jha16june]

Originally Posted by naga25french

yeah .. thats very much right...

most ppl use N *(1-1/a)*(1-1/b)...

but the euler number is basically the LCM of all (prime factors -1)numbers...

@Naga:

Well the Phi(N) in euler theorem is euler totient function and the value of phi(N) is given by n*(1-1/p)... . actually in PG Cat parlance we have given it a name of Euler Number which is not true.

Now if you take n = 45, then 45=3^2*5, so prime factors are 3 and 5, so you will calculate euler number as LCM(2, 4) which is 4. And then you will say that 2^4 when divided by 45 will give a remainder of 1, right? (2 and 45 are coprime). Which is not right.
Well I dunno if I have misunderstood ur postulate, but there is something fishy the way you have suggested to calculate the EN.

[Varun Khullar]

Originally Posted by ganesh99

[/b] can someone pls explain this

euler of prime n is n-1

1001 = 7 *11*13

now euler of 1001 might be 720 .. but
lcm of 6 ,10 ,12 is 60
so we choose it has factor

2^1001 mod 1001 =2^41 mod 1001 .. its just to reduce the term more.
2^60k mod 1001 = 1

[prakharc]

Originally Posted by jha16june

@Naga:

Well the Phi(N) in euler theorem is euler totient function and the value of phi(N) is given by n*(1-1/p)... . actually in PG Cat parlance we have given it a name of Euler Number which is not true.

Now if you take n = 45, then 45=3^2*5, so prime factors are 3 and 5, so you will calculate euler number as LCM(2, 4) which is 4. And then you will say that 2^4 when divided by 45 will give a remainder of 1, right? (2 and 45 are coprime). Which is not right.
Well I dunno if I have misunderstood ur postulate, but there is something fishy the way you have suggested to calculate the EN.

For 45 ..don't take euler number of 3 and 5....Since we have 3^2....take euler number of 9 and 5
EN(9)= 6 EN(5)=4
LCM(6,4)=12
SO 2^12 will give remainder 1 witk 45
2^12=4096 mod 45 =1

Similarly for 100= 2^2 * 5^2 ....take euler numbers of 4 ,25 and not 2,5

[sohan shetty]

wrong post.....

[bhatia.sanjay01]

Q: An examination consists of 160 questions. For each correct answer, a student earns 1 mark and for each wrong answer, a student loses 1/4th of the a mark. Find the number of distinct net scores that a student could earn in the examination.

i) 790 2) 797 3) 795 4) 801

[mahi101987]

please solve this one ........

I cut a piece of paper in four equal parts. Now, I cut exactly one out of these four parts into four equal parts. Now, again if I keep on repeating the same process for infinite number of times, then which of the following can be the number of parts of paper at any instant of time?

(1) 2048
(2) 2049
(3) 2050
(4) 2051
(5) 2052

[mahi101987]

Originally Posted by bhatia.sanjay01

Q: An examination consists of 160 questions. For each correct answer, a student earns 1 mark and for each wrong answer, a student loses 1/4th of the a mark. Find the number of distinct net scores that a student could earn in the examination.

i) 790 2) 797 3) 795 4) 801

the max possible score is 160 and the minimum possible is -40

and the difference between any scores is only 1/4 mark

ex.. 160,159*(3/4),159*(2/4),159*(1/4),159 so on..........

so like this we can get the distinct scores

[jain12345]

Originally Posted by bhatia.sanjay01

Q: An examination consists of 160 questions. For each correct answer, a student earns 1 mark and for each wrong answer, a student loses 1/4th of the a mark. Find the number of distinct net scores that a student could earn in the examination.

i) 790 2) 797 3) 795 4) 801

answer is 795
reason: you can score from -40 to 160 in steps of 1/4
eg -40,-39(1/4), .........to 160so in all 800 possibilities but u cannot score 158(1/2),159(1/4),159(1/2),159(3/4),157(1/4) thus 5 less. thus the answer is 795

cheers

[jain12345]

Originally Posted by mahi101987

please solve this one ........

I cut a piece of paper in four equal parts. Now, I cut exactly one out of these four parts into four equal parts. Now, again if I keep on repeating the same process for infinite number of times, then which of the following can be the number of parts of paper at any instant of time?

(1) 2048
(2) 2049
(3) 2050
(4) 2051
(5) 2052

the question is a bit vague as in i perceive u are talking about the number of peices since starting. if that is true the answer should be multiple of three plus one the only option being 2050

cheers