32^32^32 what would be remainder when divisible by 9
Answers
The digital root of a non-negative integer is the (single digit) value obtained by an iterative process of summing digits, on each iteration using the result from the previous iteration to compute a digit sum. The process continues until a single-digit number is reached.
For example, consider the number 32,768. For simplicity let us denote the digital root of ’n’ by D(n). So, D(32,768) = D(3+2+7+6+8) = D(26) = D(2+6) = 8.
*This can be used for verifying the result after a calculation. D(a×b)=D(a)×D(b)D(a×b)=D(a)×D(b).
Example: D(19×20)=D(380)=2D(19×20)=D(380)=2 and D(19)×D(20)=1×2=2D(19)×D(20)=1×2=2
*The remainder when a number is divided by ‘9’ is the same as it’s digital root.
Now, coming to the question: The remainder when (3232)32(3232)32 is divided by 9.
(3232)32=3232×32=321024(3232)32=3232×32=321024
=32×32×32×......1024times=32×32×32×......1024times
Now, to find the remainder, let us find D((3232)32).D((3232)32).
D((3232)32)D((3232)32)
=D(32×32×32×......1024times)=D(32×32×32×......1024times)
=D(5×5×5×......1024times)=D(5×5×5×......1024times)
=D(54×54×54×......256times)=D(54×54×54×......256times)
=D(625×625×625×......256times)=D(625×625×625×......256times)
=D(4×4×4×......256times)=D(4×4×4×......256times)
=D(44×44×44×......64times)=D(44×44×44×......64times)
=D(256×256×256×......64times)=D(256×256×256×......64times)
=D(4×4×4×......64times)=D(4×4×4×......64times)
=D(44×44×44×......16times)=D(44×44×44×......16times)
=D(256×256×256×......16times)=D(256×256×256×......16times)
=D(4×4×4×......16times)=D(4×4×4×......16times)
=D(44×44×44×44)=D(44×44×44×44)
=D(256×256×256×256)=D(256×256×256×256)
=D(4×4×4×4)=D(4×4×4×4)
=D(44)=D(44)
=D(256)=D(256)
=4=4
The remainder is 4.
Thank you. :)