32 32 32 9 give the remainder Options a 6 b 4 c 7 d 1

32^32^32=(36-4)^32^32
according to remainder theorem remainder will be (-4)^32^32
now 32^32 will be an even number so we can say that remainder will be 4^32^32
we know that 4^3=64 and 9*7=63 so we need to write 32^32 in terms of 3
so 32^32/3=(33-1)^32/3 remainder will be (-)^32=1
so we can write 32^32=3k+1
32^32 is even so 3k+1 must also be even
=>3(odd)*k(odd)+1(0dd)=even so we can say that k is odd
4^32^32=4^(3k+1)=4^3k*4
64^k*4
((63+1)^k*4)/9 so as per remainder theorem will be +1^k*4=1*4= 4 ans
10 years agoHelpfull: Yes(3) No(1)
4 is the right answer
10 years agoHelpfull: Yes(2) No(0)
unit digit=32^32=2*32=6
thus 32^6/9=2^30/9=8^10/9=(-1)^10=1 ans
10 years agoHelpfull: Yes(2) No(4)
ans is 7
=32 %9 =5
=5^ 32.32. 32. . 32 times /9
=5^ 2. 2.2.2... 32 times /9
= 7 ans
10 years agoHelpfull: Yes(0) No(3)
(32pow32pow32/9)
=(5pow1024)/9
=gives -5 as remainder
Hence 9-5=4
8 years agoHelpfull: Yes(0) No(0)

(32^32^32)/9 give the remainder..
Options
a) 6
b) 4
c) 7
d)1