2 to the power of 2 to the power of 2003

Edith, Vivien, Charlotte, Lucy, Priscilla, Francesca and Rosie all from the Mount School in York and Andrei of School 205, Bucharest all sent in correct solutions. Well done to all of you. This problem relied on you identifying the cycle in the last two digits of the solution..

You can then investigate the pattern in the powers.
Below is a table which shows the last two digits for each of the powers in a full cycle from 04 back to 04:

$\begin{matrix} 2^{1} \to 02 \\ 2^{2} \to 04 \\ 2^{3} \to 08 \\ 2^{4} \to 16 \\ 2^{5} \to 32 \\ 2^{6} \to 64 \\ 2^{7} \to 28 \\ 2^{8} \to 56 \\ 2^{9} \to 12 \\ 2^{10} \to 24 \\ 2^{11} \to 48 \\ 2^{12} \to 96 \\ 2^{13} \to 92 \\ 2^{14} \to 84 \\ 2^{15} \to 68 \\ 2^{16} \to 36 \\ 2^{17} \to 72 \\ 2^{18} \to 44 \\ 2^{19} \to 88 \\ 2^{20} \to 76 \\ 2^{21} \to 52 \\ 2^{22} \to 04 \end{matrix}$

From the table you can see that the last two digits of powers of 2 go in cycles of 20 and so

2^{2003}

will end in the two digits 08. We only need to look at the last two digits in the final stage of the solution because we are always looking at cycles of 20 - one digit is not enough (it would only enable us to cover all possibilities in a cycle of ten or less) and three digits is more than we need.

But we need to raise 2 to the power

2^{2003}

and identify what happens to the last digits in this case.

As the pattern of the last two digits will also be based on the same cycle we can use the table to see what the last two digits will be for a power ending 08 (which is 8 mod 20) and this is 56.