Bunuel wrote:

The last digit of 3 in positive integer power repeats in pattern of 4: {3, 9, 7, 1}. So, basically we should find the remainder upon division 7^(11) by cyclicity of 4 (to see on which number in this pattern \(7^{11}\) falls on). \(7^{11}=(4+3)^{11}\), now if we expand this expression all terms but the last one will have 4 in them, thus will leave no remainder upon division by 4, the last term will be \(3^{11}\). Thus the question becomes: what is the remainder upon division \(3^{11}\) by 4:

Hi Bunuel,

I didn't understand why we are now trying to focus on \(3^{11}\) by 4....

We stopped looking at the rest of the question and just looking at the exponents... why is this rigth?