what is the remainder of ((32)^32)^32/7 ?please give the logic applied .........
i find really hard to answer such question having power thrice times.thanx
what is the remainder of ((32)^32)^32/7 ?please give the logic applied .........
i find really hard to answer such question having power thrice times.thanx
i think the answer is 4
method:
32=4(mod7)
so, ((32)^32)^32(mod7)=(4^(32*32))(mod7) =4^1024(mod7)
4^1(mod7)=4;
4^2(mod7)=2;
4^3(mod7)=1 and the pattern repeats. so the period is 3
so, 1024=3*341+1
so answer is 4(mod7)
am i right?
Hey guys i got stuck with this Question..help me out..
What is the 28383rd term of the series 12345678910111213.....
Regards,
Ces
i think the answer is 4
method:
32=4(mod7)
so, ((32)^32)^32(mod7)=(4^(32*32))(mod7) =4^1024(mod7)
4^1(mod7)=4;
4^2(mod7)=2;
4^3(mod7)=1 and the pattern repeats. so the period is 3
so, 1024=3*341+1
so answer is 4(mod7)
am i right? :lookround:
Yeah u can use eulers method to solve the same..Its very simple and fast to solve such remainder problems..
Eulers theorem---
How to find remainder for a^f(M)/N
Here a and N must be coprimes to use this method
Here f(M)=Nos of no less than N and coprime to it
for eg f(10)={1,3,7,9} ,4 coprimes are there less than 4
f(20)={1,3,7,9,11,13,17,19}..8 coprimes are there less than 20
Now for highes values there is a formulae which we can use to find this number of coprimes
Express M as product of coprimes
M=a^x*b^y*c^z...(where a,b,c are coprimes)
f(M)=M(1-1/a)(1-1/b)(1-1/c)
eg if M is 63 then f(63)=3^2*7
therefore f(63)=63(1-1/3)(1-1/7)=36
which means there are 36 coprimes less than 63..
eg...find remainder for 53287^1374285/20
How to go about it
1st step>find the number of coprimes less than 20
now 20=2^2*5
by using our formulae
f(M)=20(1-1/2)(1-1/5)=8
now divide the power by 8..
using divisibilty rule of 8 here u have to only look at the last three digits i.e 285..the remainder will be 5
so the earlier expression boils down to 7^5/20
which is simple now to solve
i.e 7^2*7^2*7/20
=49*49*7/20
=9*9*7/20
=81*7/20
=1*7/20
which means here the remainder is 7..
Here I have explained each and every steps in detail...to make it lucid
Now lets try to solve using this method the original question asked
(32)^32)^32/7
Here the euler number for 7 i.e the number of co primes less than 7 is 6
so divide the outer power 32 by 6 which u will get remainder as 2
so now the question becomes
Find the power for
(32^(32))^2/7
which is
32^64/7
Now again diving 64 by 6 we get remainder as 4
32^4/7
also now divde 32 by 7 to get remainder as 4
therefore now the question boils down to
4^4/7
which 16*16/7
and here the remainder is 4..
Hope u understood the method bro...
along a long corridor there are 100 doors marked as 1,2,3....100. as you know the doors can be in two states open or close. initially all doors are closed. person number 1 changes the status of doors which are multiple of 1 i.e. all doors , then person 2 changes the state of all doors which are multiple of 2 and person 3 change the state of doors which are multiple of 3 and so on till person 100.
now how many doors are closed ?
pls provide solution for this !
along a long corridor there are 100 doors marked as 1,2,3....100. as you know the doors can be in two states open or close. initially all doors are closed. person number 1 changes the status of doors which are multiple of 1 i.e. all doors , then person 2 changes the state of all doors which are multiple of 2 and person 3 change the state of doors which are multiple of 3 and so on till person 100.
now how many doors are closed ?
pls provide solution for this !
Is the answer 70?
Unruffled SaysIs the answer 70?
answer is nor available to me... :(
but the hint given is that, try to find out for a particular door how many persons will change the state!
answer is nor available to me... :(
but the hint given is that, try to find out for a particular door how many persons will change the state!
it is a standard type of question.
in such kind of situations always the doors whose nos. are perfect squares will remain open the rest will be closed, so 10 doors will be open, 90 will be closed.
i hope you get it.:cheerio:
it is a standard type of question.
in such kind of situations always the doors whose nos. are perfect squares will remain open the rest will be closed, so 10 doors will be open, 90 will be closed.
i hope you get it.:cheerio:
Thanks ! but using ur logic shouldn't the answer be 91, as there are 9 perfect squares till 100, which will get deducted from the no. of doors. ?
My logic says... prime number doors+ door numbers which have even factors till 100 = answer ; 25 + 66 = 91
wt sy ?
i think you are forgetting that 1 is a square no. too! 😃
madtwin Saysi think you are forgetting that 1 is a square no. too! :)
Got it
thanks !
@ces bro but from the last you have written (32^(32))^2/7 as 32^64 but it should be (32)^32*32/7
than 6 divides 32*32 than get 4 as remainder than 32^4/7.
one more question why we are first dividing 32/7 without any care of power. or better we can take the trend of remainder.
like 32^1/7=4; 32^2/7=2; 32^3/7=1;32^4/7=4means cycle of three...........and 32 is one more of three
so remainder is 4.
@ces bro but from the last you have written (32^(32))^2/7 as 32^64 but it should be (32)^32*32/7
than 6 divides 32*32 than get 4 as remainder than 32^4/7.
one more question why we are first dividing 32/7 without any care of power. or better we can take the trend of remainder.
like 32^1/7=4; 32^2/7=2; 32^3/7=1;32^4/7=4means cycle of three...........and 32 is one more of three
so remainder is 4.
Originally Posted by jain_ashu View Post
If its 1!+2!+...+1000! then the ten's digit is 0.From 10! onwards no number contributes to 10's digit.
If its 5!+6!+7!+.....+1000! then the ten's digit is 8. 
didnt able to understand ye 8 kaise aya!!!!
Originally Posted by palakp View Post
along a long corridor there are 100 doors marked as 1,2,3....100. as you know the doors can be in two states open or close. initially all doors are closed. person number 1 changes the status of doors which are multiple of 1 i.e. all doors , then person 2 changes the state of all doors which are multiple of 2 and person 3 change the state of doors which are multiple of 3 and so on till person 100.
now how many doors are closed ?
pls provide solution for this !
see answer is 10 as there are 10 perfect square and only perfect square has odd no factors there for only those doors which have odd no of factors will be open and rest will be closed that is 90
now puys answer this in the above question if there are 70 ppl not 100 then how many doors will remain closed!!!(rest of the condition remains same)
Originally Posted by jain_ashu View Post
If its 1!+2!+...+1000! then the ten's digit is 0.From 10! onwards no number contributes to 10's digit.
If its 5!+6!+7!+.....+1000! then the ten's digit is 8.
didnt able to understand ye 8 kaise aya!!!!
see 5! Wil have its tens digit as 2 ,6! Wil have 2...similarly find until 9! N add dem coz after dat v vil have 2 zeroes at the end...the sum at tens digit comes out to be 8.
ashishkakkar123 Saysnow puys answer this in the above question if there are 70 ppl not 100 then how many doors will remain closed!!!(rest of the condition remains same)
the answer then will be 8 I guess...
find the no. of perfect square factors of 3600? also explain.
rohitaryan Saysfind the no. of perfect square factors of 3600? also explain.
3600 = 4 x 9 x 100 = 2^2 x 3^2 x 10^2
So, ans is 3