# Number System - Questions & Discussions

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. H…

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/discussions/number-system-25000852

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

http://www.pagalguy.com/discussions/conceptstotal-fundas-25023536

the official Quant thread for unsorted queries:

Hope that helps.

All the best puys

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
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.
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?
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/-
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
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
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

option II only

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

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

janvats Says
is t answer c only II?

naga25french Says
c) only II

Answer is 28 only but prove it guys.
find the remainder when (38^16!)^1777 is divided by 17

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

E(17) = 16
(38^16!)^1777 mod 17=
38^16! mod 17 = 1
remainder = 1
find the remainder when (38^16!)^1777 is divided by 17

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

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

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

Answer should be 1.
E(17) = 16, (38^16!)^1777 mod 17 = 38^16k mod 17 = 1

yes the ans is one

will try to be regular from now onwards..(in this thrd)

okay one qstn from my side

5789^(1!+2!+3!-------+1000!)has the unit digit of ?

yes the ans is one

will try to be regular from now onwards..(in this thrd)

okay one qstn from my side

5789^(1!+2!+3!-------+1000!)has the unit digit of ?

Hi Pk,

As the cyclic form will give only two possibility as 9 or 1 as last digit depending on the power (if odd then 9, if even then 1)

So let us work on power. After the 5! all the sum will have last digit as 0, so we have to work only upto 4!
1!=1, 2!=2, 3!=6, 4!=24 sum of all will give 33. So we are concerned only with 3 as last digit of power which is ODD. So the answer will be 9.
yes the ans is one

will try to be regular from now onwards..(in this thrd)

okay one qstn from my side

5789^(1!+2!+3!-------+1000!)has the unit digit of ?

the answer is 9

as 1!+ 2! + 3! +4! ends in 3

9^odd gives 9 as unit digit
yes the ans is one

will try to be regular from now onwards..(in this thrd)

okay one qstn from my side

5789^(1!+2!+3!-------+1000!)has the unit digit of ?

1+2+6+24+20+20+40+20+80 = xx13
5789^xx13
unit digit is 9
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

we can write expression as 2^(4n+1)-2^2n
Last digits of 2^(4n+1) are in cycle 32,92,52,12,72
Last digits of 2^2n are in cycle 04,64,24,84,44
Last 2 digits are always 28