[Mani_23]

Originally Posted by shanks4mba

1)is it 62^62^62? in the question

2) (17) (9!) + 18! mod (9!) 8704.
now 17*9!mod9!*8704 = 1mod512 = 1*9!*17 = 9!*17
18!mod9!*17*512 = 18*17*16*15*14*13*12*11*10mod17*2^9
now 18*17*16*15*14*13*12*11*10mod17 = 0
and 18*17*16*15*14*13*12*11*10mod2^9 = 2^4(9*17*2^3*15*7*13*6*11*5)mod2^9
2^8(9*17*15*7*13*3*11*5)mod2^9 = 1*2^8
hence 2^9k + 2^8 = 17m
remainder = 2^12 + 2^8 = 2^8(17) * 9!
hence total remainder = 2^8(17) * 9! +9!*17
= 17*9!(1+256) = 257*9!*17???

no..for the second ques..ans is 17*9!..

[shanks4mba]

Originally Posted by Mani_23

no..for the second ques..ans is 17*9!..

srry did a mistake here instead of 2^4 it will be 2^5 hence the remainder for the second part will be 0
hence the total remainder = 17*9!.

will solve the first question later...u can use euler's theorem in there./just divide by 1000

[hiashish19]

Originally Posted by shanks4mba

1)is it 62^62^62? in the question

2) (17) (9!) + 18! mod (9!) 8704.
now 17*9!mod9!*8704 = 1mod512 = 1*9!*17 = 9!*17
18!mod9!*17*512 = 18*17*16*15*14*13*12*11*10mod17*2^9
now 18*17*16*15*14*13*12*11*10mod17 = 0
and 18*17*16*15*14*13*12*11*10mod2^9 = 2^5(9*17*2^3*15*7*13*6*11*5)mod2^9
2^8(9*17*15*7*13*3*11*5)mod2^9 = 0
hence total remainder = 9!*17 + 0
=9!*17

18! = 2*9*17*(2^4)*15*2*7*(2^2)*3*11*2*5*9!
= 2^(1+4+1+2+1)*17*9!*k
= 2^9*17*9!*k
= 8704*9!*k

divisible

[hi.shivani]

Another one:

N is a three digit no. written in base 10. N1 is the same no. written in base 9 and N2 in base11. Whats is the probability that N1 and N2 are also three digit numbers ?

[hiashish19]

Originally Posted by hi.shivani

Another one:

N is a three digit no. written in base 10. N1 is the same no. written in base 9 and N2 in base11. Whats is the probability that N1 and N2 are also three digit numbers ?

nt sure may be 608/900

(88 in base 9 = 728
100 in base 11 = 121

728-121 +1 = 608 nos. possible

[jaidev_iyer]

Originally Posted by hi.shivani

Another one:

N is a three digit no. written in base 10. N1 is the same no. written in base 9 and N2 in base11. Whats is the probability that N1 and N2 are also three digit numbers ?

Is the answer 609/900?

[Ankurb]

Originally Posted by hi.shivani

Another one:

N is a three digit no. written in base 10. N1 is the same no. written in base 9 and N2 in base11. Whats is the probability that N1 and N2 are also three digit numbers ?

range:-
121 <= N <= 728
probability = 608/900

[Ankurb]

Originally Posted by Mani_23

1.find the last three digits of 62^62^62

1.
62^62^62 mod 2^3 = 0
62^62^62 mod 5^3 =
62^(100k+44) mod 5^3 =
2^44 mod 5^3 = 16^11 mod 5^3 = 1 +15*11 mod 5^3 = 41
31^44 mod 5^3 = 1+44*30 + 22*43*900 mod 5^3 = 96
remainder = 96+41 = 137 - 125 = 12
125k + 12 = 8p, ==>k = 4
last 3 digits are :- 512

[hi.shivani]

Originally Posted by Ankurb

range:-
121 <= N <= 728
probability = 608/900

Elaborate plzz..!

[hiashish19]

Originally Posted by Ankurb

range:-
121 <= N <= 728
probability = 608/900

got mistake .. ignore....