2349 = 1 (mod 4)
2^(4n+1) = 2 (mod 10)
Therefore last digit of 2^2349 is 2
In fact, 2^2349 = 131661808029361064474122541992202249544316172470474290896
938957889657411706431738059855987100606116738608709863898
114516322696806682779725395040361126073989754846342223650
752839934282507553095228819445591564242224229100999549808
158043220482537489286899542806927109262299798576702052364
647531326323060875437512721666750074735365806697524459192
563914773873444570420147733289826848369892651509013842153
766576752759546709772991670483710813378588223114875800405
103605361303657861719088971021114521028156816509308794659
625756698728308129465992649969171244954433589856461308546
917926791439305402526271547340075851639933262085702464422
986758336147589425826676630311810520418347644992379850195
840502843574915602317312
😛
Originally posted by THUDandBLUNDER😲Wow. I'm impressed. Any calculator/spreadsheet I've ever used starts throwing up after about 10 number. How did you get all those???
2349 = 1 (mod 4)
2^(4n+1) = 2 (mod 10)
Therefore last digit of 2^2349 is 2
In fact, 2^2349 = 131661808029361064474122541992202249544316172470474290896
938957889657411706431738059855987100606116738608709863898
114516322696806682779725395040361126073989754846342223650
752839934282507553095228819445591564242224229100999549808
1580432 ...[text shortened]... 2
986758336147589425826676630311810520418347644992379850195
840502843574915602317312
😛