What is the remainder when 32^32^32 is divided by 7?

VeritasKarishma

Many Thanks for your explanation. Can you please help me with the question posted by Gurpreet, as one more example will help me get hold of the logic.

Remainder when 32^32^32 is divided by 9.

My solution,

We need to only think about 2^32^32

And, we know that 2^3 = 8 which is 9-1. So again we want to know can we write 32^32 as 3k+1.

=> (33-1)^32, now as the exponent is even we will have remainder as 1.
=> 2^32^32 = 2^(3k+1)
=> 2 x 2^3k = 2 x 8^k
=> 2 x (9-1)^k.

This is where I get stuck because I don't know whether k is even or odd.

If k is even, then we have 2 divided by 9
remainder = 2

and, if k is odd,

we have -1 x 2 = -2

and remainder = 9-2 = 7.

Can you please help me in this.

I tried doing this by (27+5)^32^32

and there too I get (126-1)^k and it again depends on k being even or odd.

Edit:

I thought about it and did the question as below, please let me know if the methodology is correct.

Remainder when 32^32^32 is divided by 9?

32^32^32 = (36-4)^32^32

Now, 36^x will always be divisible by 9 so we need to look at only 4^32^32

We also know that 4^3 = 64, which is 63+1. So we need to find whether 3 will leave any remainder in 32^32.

(33-1)^32/3 leaves a remainder of 1.

Therefore, we can write 4^32^32 as 4^(3k +1)

= 4 x 4^3k = 4 x 64^k
= 4 x (63+1)^k
= 4

Thus 32^32^32 leaves a remainder of 4 when divided by 9.

Is my solution correct?


Thanks
Saurabh