What is the remainder when 3 ^{2001} is divided by
7?

Zi Heng Lim and Hagar El Bishlawi found a pattern when they raised 3 to a power, divided it by seven and found the remainder.

3 ^{1} = 33 ^{2} = 93 ^{3} = 273 ^{4} = 813 ^{5} = 2433 ^{6} = 729--Pattern Found-- |
3/7=0 R3 9/7=1 R2 27/7=3 R6 81/7=11 R4 243/7=34 R5 729/7=104 R1 |

3 ^{7} = 21873 ^{8} = 65613 ^{9} = 196833 ^{10} = 590493 ^{11} = 1771473 ^{12} = 531441 |
2187/7=312 R3 6561/7=937 R2 19683/7=2811 R6 59049/7=8435 R4 177147/7=25306 R5 531441/7=75920 R1 |

When you divide the exponent 2001 by the number 6, the answer
will be 333 R3. The third number in the remainder pattern is 6,
therefore 3 ^{2001} divided by 7 has a remainder of 6,
assuming that the pattern keeps repeating itself every six
powers.

Can you prove that this pattern will keep repeating itself?

Pierce Geoghegan and Etienne Chan solved this problem using modulus arithmetic. This is Pierce's solution:

Initially we use the fact that 3 ^{n} (mod 7) =
3 x [3 ^{n-1} (mod 7)]

so

3 ^{1} = 3(mod 7) = 3

3 ^{2} = 3 x 3(mod 7) = 2

3 ^{3} = 3 x 2(mod 7) = 6

3 ^{4} = 3 x 6(mod 7) = 4

3 ^{5} = 3 x 4(mod 7) =5

3 ^{6} = 3 x 5(mod 7) = 1

Now we have 3 ^{6} = 1 mod 7 and it follows that 3
^{1998} = (3 ^{6} ) ^{333} = 1
^{333} (mod 7) = 1

But 3 ^{2001} = 3 ^{1998} x 3 ^{3} = 1 x
3 ^{3} (mod 7) = 6

therefore the remainder is 6 when 3 ^{2001} is divided by
7.