Number System - Questions & Discussions

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
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

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)
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)

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

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.
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
Q1)Let N be the product of five different odd prime numbers. If N is the five-digit number abcab, 4
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
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 ???
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.
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
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
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
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???
Q1)Let N be the product of five different odd prime numbers. If N is the five-digit number abcab, 4
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
divishth Says
Is it 96 years ???

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.


Normal year 52 weeks + 1 day(84/7)
Leap year 52 weeks + 2 days(84/4*7)

/365 = 12 years and 3 days, hence 96 years

kindly let me know, if you have any observation..

Cool thread..!!

I will posting my doubts and any new thing that i will be coing across..!!

Lets contribute.

guys hav a question here: Progressions.

If a times ath term of an AP equals b times bth term. find the (a+b)th term.

- What i wanna ask is, is there a non-algebraic way of cracking this? A short-cut... u go thru the formulae, things work out fine.. but takes a lil time...

Thanks much!

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


1)N = abcab where 4 N has to be in the form 1001* P1*P2
where 1001 = 7*11*13
N = 1001*P1*P2
as 4 so we have set p1,p2 = (3*17) , (3*19) and (3*23)
hence total 3 values of N are possible
option c
guys hav a question here: Progressions.

If a times ath term of an AP equals b times bth term. find the (a+b)th term.

- What i wanna ask is, is there a non-algebraic way of cracking this? A short-cut... u go thru the formulae, things work out fine.. but takes a lil time...

Thanks much!

TAKE An A.P = 2 1 0 -1...
let a = 1 and b = 2
the case is satisfied...so a+bth term is the 3rd term which is 0:)

Another approach...
given a(x + (a-1)d) = b(x + (b-1)d)
take a = 1 and b = 2 then
x = 2x + 2d, assume d = -1
then x = - 2d = 2
so u get A.P = 2,1 0 -1... hence the third term is 0:)

PS-always try to assume values when it comes to Progressions, thn u can find out the ans quickly. most of the time this works:). Same wit Time and work too:)