What is the remainder when 9^1995 is divided by 7?

Note that :

9^1 mod 7 = 2

9^2 mod 7 = 4

9^3 mod 7 = 1

In general, for any integer n ≥ 0,

9^(1 + 3n) mod 7 = 2

9^(2 + 3n) mod 7 = 4

9^(3 + 3n)mod 7 = 1    

And 1995 = 3 + 3(664)

So.....9^1995 mod 7  = 1

What is the smallest positive integer n such that n has remainder 1 when divided by 3, n has remainder 1 when divided by 4, and n has remainder 4 when divided by 5?

n = 60 m+49, m is any positive integer.

n = 60.1 +49 =109 is the smallest positive integer that satisfies all above conditions.

CORRECTION: In the above answer I meant to multiply (60 x 0) + 49 =49 would be the smallest positive integer.

2)Using CPhill's method......

N = 3x + 1 ----(1)

N = 4y + 1 ----(2)

N = 5z + 4 ----(3)

(1) x 2 - (2) - (3)

6x + 2 - (4y + 1) - (5z + 4) = 0

6x + 2 - 4y - 1 - 5z - 4 = 0

6x - 4y - 5z = 3

Smallest integer solutions are: x = 16, y = 12, z = 9

N = 3(16) + 1 = 49

Final answer:49!!