[shashank3012]

Hello people,

As the earlier number system thread had more than 10k replies,we have closed the old one and presenting to you a part 2 of the same

Please continue with the discussions of Number system problems here,henceforth.

Here is the link to the old thread:

http://www.pagalguy.com/forum/quanti...er-system.html (Number System)

Also,for getting the concepts,here is a good thread for the same:

http://www.pagalguy.com/forum/quanti...al-fundas.html (Concepts...total fundas!!)

the official Quant thread for unsorted queries:

http://www.pagalguy.com/forum/quanti...ml#post1605026 (Official Quant thread for CAT 09-II [July 09 onwards])

Hope that helps.

All the best puys

[Ankurb]

Let's start things afresh
Q
If n is an odd integer then the last two digits in base ten of 2^2n * (2^(2n+1) -1) are
I. 18 II. 28 c. 38
a. Only I
b. Only I & II
c. Only II
d. Only I, II & III
e. None of the above

[jain_ashu]

1. 3^32/50 give remainder and {.}denotes the fractional part.the fractional part is of the form(0.bx) value of x is??

3^32 mod 50 = 41
So,bx = 82 => x = 2.

[janvats]

Originally Posted by Ankurb

Let's start things afresh
Q
If n is an odd integer then the last two digits in base ten of 2^2n * (2^(2n+1) -1) are
I. 18 II. 28 c. 38
a. Only I
b. Only I & II
c. Only II
d. Only I, II & III
e. None of the above

is t answer c only II?

[jain_ashu]

Originally Posted by Ankurb

Let's start things afresh
Q
If n is an odd integer then the last two digits in base ten of 2^2n * (2^(2n+1) -1) are
I. 18 II. 28 c. 38
a. Only I
b. Only I & II
c. Only II
d. Only I, II & III
e. None of the above

Is the answer option c) Only II ??
And the answer to the question you had earlier posted is Rs.50/-

[naga25french]

Originally Posted by Ankurb

Let's start things afresh
Q
If n is an odd integer then the last two digits in base ten of 2^2n * (2^(2n+1) -1) are
I. 18 II. 28 c. 38
a. Only I
b. Only I & II
c. Only II
d. Only I, II & III
e. None of the above

c) only II

[vyomconfused]

Originally Posted by Ankurb

Let's start things afresh
Q
If n is an odd integer then the last two digits in base ten of 2^2n * (2^(2n+1) -1) are
I. 18 II. 28 c. 38
a. Only I
b. Only I & II
c. Only II
d. Only I, II & III
e. None of the above

answer is c only i.e 28

[shanks4mba]

Originally Posted by Ankurb

Let's start things afresh
Q
If n is an odd integer then the last two digits in base ten of 2^2n * (2^(2n+1) -1) are
I. 18 II. 28 c. 38
a. Only I
b. Only I & II
c. Only II
d. Only I, II & III
e. None of the above

take n = 1
then we have 2^2n * (2^(2n+1) -1) = 4*(7) = 28
take n = 3
then we have 64*127 = 28
basically its even*odd = (x4)*(y7) = x28
shud be opption C only II

[sachy24]

option II only

[janvats]

find the remainder when (38^16!)^1777 is divided by 17

a. 1
b. 16
c. 8
d. 13