Q1)Let N be the product of five different odd prime numbers. If N is the five-digit number abcab, 4 < a < 8, how many values of N are possible?
a. 1
b. 2
c. 3
d. 4
e. more than 4

option c) 3
abcab
there is a pattern there
3*7*11*13 = 3003
3003*17 = 51051
3003*19 = 57057
3003*23 = 69069

as 4<8, we stop here

for grandfather 7 days=6 days..

so no. of yrs=84*7/6=98

In 84 years he is missing out almost 84*52 Sundays plus max. 21 for leap years. Which is equal to 4368+21 = 4389 = approx. 12 years.
Where are we going wrong???

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

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

Is the ans 1?
i think we can do it by Euler's..plz comment

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

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

38^16! mod 17 = 4^16! mod 17 = (4^2)^even no mod 17 = (-1)^even no mod 17 = 1
thus 1^ 17777 mod 17 = 1

jain_ashu Says

Don't have the OA but I am too getting the same answer. But would like to know your approach especially regarding leap year part.

for grandfather 7 days=6 days..

so no. of yrs=84*7/6=98

divishth Says

Is it 96 years ???

Don't have the OA but I am too getting the same answer. But would like to know your approach especially regarding leap year part.

Q1)
Q2) My grandfather said he was 84 years old but he was not counting the Sundays. How old my grandfather really was?
a. 96
b. 97
c. 98
d. 64

Is it 96 years ???

Q1)Let N be the product of five different odd prime numbers. If N is the five-digit number abcab, 4 < a < 8, how many values of N are possible?
a. 1
b. 2
c. 3
d. 4
e. more than 4

Q2) My grandfather said he was 84 years old but he was not counting the Sundays. How old my grandfather really was?
a. 96
b. 97
c. 98
d. 64

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 ?

after 5!,,each term will end with 0..so we'll add 1!+2!+3!+4!= 1+2+6+24=33

and last two digits of 5!+6!-----9!=80..adding 80+33=13..

5789^13/10
(5780+9)^13/10
9^13/10=9 unit digit

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

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

ans:1

(38^16!)^1777=(34+4)^16!^1777/17

4^16!^1777/17

16! is divisbile by 4 so (4^2)4k^1777/17

(16+1)^4k^1777/17..

so 1 remainder.

this is fermat little theorem
this can be seen rem(M^(N-1)/N) = 1 where N is a prime number
so the answer is 1*1*...............1 =

sorry for the wrong explanation
I studied post wrong
will solve and get back 2 u

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

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

(38^16!)^1777mod 17 is 1
option a)

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 ?

unit digit is 9 as the power is odd ( coz except 1! all other factorials are even)

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 ?

last digit is 9 as power is odd.

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

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

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

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 ?

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

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 ?

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

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

E(17) = 16
(38^16!)^1777 mod 17=
38^16! mod 17 = 1
remainder = 1

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

option II only

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

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

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

is t answer c only II?

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