[Ankurb]

Originally Posted by janvats

is t answer c only II?

Originally Posted by naga25french

c) only II

Answer is 28 only but prove it guys.

[Ankurb]

Originally Posted by janvats

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

[sachy24]

Originally Posted by janvats

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

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

the ans is one

[jain_ashu]

Originally Posted by janvats

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

[pk_gt1]

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 ?

[TAM_Pain]

Originally Posted by pk_gt1

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.
Please confirm

[naga25french]

Originally Posted by pk_gt1

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

[Ankurb]

Originally Posted by pk_gt1

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

[aryan007]

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

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

[vyomconfused]

Originally Posted by pk_gt1

yes the ans is one



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

1! +2!+3!+4! = 33
so, 9^33= 9
hence, answer is 9