Find the remainder when 39! is divided by 41.
Ans is 1. but please give the full explanation.
Thanks in advance.
chiaotzu SaysThanks for the reply man. Yes, I'm aware of the formula and usage of 'K'. But isn't that method trial and error? Ain't there a definite method to find out the answer, especially when the number is large and we need to specifically check if its divisible by some number with hardest possible divisibility rule. :P
Not any that I have heard of..
Find the remainder when 39! is divided by 41.
Ans is 1. but please give the full explanation.
Thanks in advance.
This problem is the application of Wilson's Theorem
According to Wilson's Theorem Rem = 1
So there u go, when p=41, (41-2)!/41=1
Not any that I have heard of..
This problem is the application of Wilson's Theorem
According to Wilson's Theorem Rem = 1
So there u go, when p=41, (41-2)!/41=1
thanks for reply but I think that Wilson's Theorem is
Rem = 1 ,
so using this i am able to solve for 40! but not for 39!
correct me if i m wrong.
thanks for reply but I think that Wilson's Theorem is
Rem = 1 ,
so using this i am able to solve for 40! but not for 39!
correct me if i m wrong.
No wilson theorem is (p-2)!/p yaar
lets take couple of simple ex to prove the result
consider p=5
so 3!/5=>6/5=1
p=7
so 5!/7=120/7=1
the formula ur taking about is (p-1)!+1 is divisible by p
try to solve this problem I will show the application of above theoremtry to solve this problem
find remainder of 23!/47..
No wilson theorem is (p-2)!/p yaar
lets take couple of simple ex to prove the result
consider p=5
so 3!/5=>6/5=1
p=7
so 5!/7=120/7=1
bro,i think there is some error in it.
Just put p=6,8,9
bro,i think there is some error in it.
Just put p=6,8,9
yaar I meant p=prime numbers... and offcourse it wouldn`t work for 6,8 and 9 as they are not prime numbers...
EagleMenace Saysyaar I meant p=prime numbers... and offcourse it wouldn`t work for 6,8 and 9 as they are not prime numbers...
eagle bhai is correct...
approach using wilson's theorem--as stated by eagle bhai...
rem-40! mod 41=40
so,40!=41*(constant say P) +40
==> 40*39!=41P+40
now,41p +40 gives rem 40 with 41 ,so lhs shud also give the same result
so rem=1
hope m clear ...
Along a road lie an odd number of stones placed at intervals of 10 m.These stones have to be assembled around the middle stone .A person can carry only one stone at a time. A man carried out the job starting with stone in middle carrying stones in succession thereby covering a distance =4.8 km.no of stones is=
a.35 b.15 c.29 d.31
ans=d
Lets take simple one and understand. Let a1, a2 , c, b1 & b2 be 5 stones.
Now, c is the the middle stone and a1, a2 lie on the LHS of c and b1,b2 lie on RHS of c.
distance b/n c-b1 = 10m
distance b/n b1-b2 = 10 m, so distance b/n c-b2 = 20 m
So to carry all the stones on the RHS(b1,b2) of c to c will be 20(c-b1 to and fro distance) + 40 (c1-b2 to and fro distance) = 60 m
RHS distance to be travelled will be like 20+ 40 + 60 +.... (AP with common difference as 20)
Similarly for LHS distance to be travelled will be 60 m. Total = 120 m.
for our case, as total distance is 4800m , the distance man travels to carry all the stones available on either side will be 2400m.
so, 20+40+60+.... = 2400
20(n(n+1)/2) = 2400
n =15 => number of stones on right hand side
So total stones = 15+1+15 = 31
Find the remainder when 39! is divided by 41.
Ans is 1. but please give the full explanation.
Thanks in advance.
------------------------------------------------------------
Wilson's says (p-1)!/p gives remainder p-1 and (p-2)!/p gives remainder 1...!!!
a doubt...
the power of 45 that will exactly divide 123!
now i know a procedure to solve this....
45 can be expressed as 3^2*5....
so i checked (123/5)+(123/5^2)...the answer comes out to be 28.....
but the answer given is 31....
please explain...where am i going wrong....
a doubt...
the power of 45 that will exactly divide 123!
now i know a procedure to solve this....
45 can be expressed as 3^2*5....
so i checked (123/5)+(123/5^2)...the answer comes out to be 28.....
but the answer given is 31....
please explain...where am i going wrong....
dear,
u r rt..............
a doubt...
the power of 45 that will exactly divide 123!
now i know a procedure to solve this....
45 can be expressed as 3^2*5....
so i checked (123/5)+(123/5^2)...the answer comes out to be 28.....
but the answer given is 31....
please explain...where am i going wrong....
Even i did the same question yesterday from Arun Sharma, the answer given is wrong. It has to be 28 no doubt about that.
hey bro mera to 32 ans aa raha hai.......
(123/5)+ (24/5) + (123/25) = 24+4+4=32
there are many errors in arun sharma book!! if u feel confident about ur answer move ahead, dont break ur head and waste time tat u got a wrong answer!!
hey bro mera to 32 ans aa raha hai.......
(123/5)+ (24/5) + (123/25) = 24+4+4=32
Eliminate the bold part.. remaining is your answer = 28
in a big hostel,there are 1000 rooms.in that hostel only even numbers are used for door numbers i.e. the rooms are numbered 2,4,6,.....2000.all the rooms have one resident only.one morning,the warden calls all the residents and tells them to go back to their rooms as well as multiples of their rooms numbers.when a guy visits a room and finds the door open,he closes it,and if the door is closed,he opens it.all 1000 guys do this operation.all the doors were opened initially.
find the total no. of doors that are closed at the end of operation?
what two digit number is less than the sum of the square of its digits by 11 and exceeds their doubled product by 5 ........................... i dont know why am i not getting the answer..... its such a simple question ... arun sharma
vijeta.jena Sayswhat two digit number is less than the sum of the square of its digits by 11 and exceeds their doubled product by 5 ........................... i dont know why am i not getting the answer..... its such a simple question ... arun sharma
sory tats nt the ans
vijeta.jena Sayswhat two digit number is less than the sum of the square of its digits by 11 and exceeds their doubled product by 5 ........................... i dont know why am i not getting the answer..... its such a simple question ... arun sharma
number=10x+y
now 10x+y=x^2+y^2-11
and 10x+y=2xy +5
so, x^2+y^2-11=2xy +5
x^2+y^2-2xy=16
x-y=4,-4
now check the options...options here are confusing
15,95 should be the answer...while 2nd option is again telling 95
:-Pdon't get confused...