Remainder of googolplex divided by 7?

googleplex is a power of ten, and 10 mod 7 is 3. Thus, mod 7, googolplex (which is 10^(10^100) is the same as 3^(10^100) mod 7. If you work out the powers of 3 mod 7, you get

3^2 == 2

3^3 == 3*2 == 6 (same as -1, incidentally)

3^3 == 18 == 4

3^4 == 12 == 5

3^5 == 15 == 1

So if you list the powers of 10 mod 7 you get

3

2

6

4

5

1

3

2

6

4

5

1

3

2

6

4

5

1

.

.

.

with every sixth one congruent to 1. So now the question is, what is the remainder when 10^100 is divided by 6? Here it is much simpler--

10 == 4

10^2 == 40 == 4

10^3 == 40 == 4

and so on; every power of 10 is congruent to 4 mod 6. So in the list of powers of 10 mod 7, the last exponent divisible by 6 (which produces 1 in the list) is just 4 short of 10^100. Counting down in the list 4 more steps gives you 4. Thus the remainder when you divide googolplex by 7 has to be 4.

The remainder is 4:

10 ≡ 3 (mod 7), then 10^(10^100) ≡ 3^(10^100) (mod 7) and

for n ≡ 0, 1, 2, 3, 4, 5 (mod 6) we have correspondingly:

3^n ≡ 1, 3, 2, -1, -3, -2 (mod 7), so the question is now: what is the remainder of 10^100 by modulo 6?

10^100 ≡ 4^100 ≡ 2^200 (mod 6). Next for

m = 1, 2, 3, 4, 5,...

2^m ≡ 2, 4, 2, 4,... (mod 6). So, 200, being an even number produces

2^200 ≡ 4 (mod 6), that means

10^(10^100) ≡ 3^4 ≡ -3 ≡ 4 (mod 7)

.428571428571...., or merely a remainder of 3.

Any integer ending in 0 (other than 0 itself), when divided by seven, results in a remainder of the above.

At least I hope that's right.

UPDATE: Man, I gotta get some coffee. You're absolutely correct.