Show that the remainder when 2^1990 (2 to the power of 1990) is divided by 1990 equals 1024.

Using extended precision arithmetic would be brute force in the strongest sense: 2^1990 =
1121221382103764183821126173025080062521794630945991406716447985181412364640986
63462564402960596315421723238734276119411469445211021782747209563609000649725135
86913002471902343817263588695316881975481151373387328591954635121717613666730178
32426429237997119107013304675868599898492415152827501603730179030091251649501801
35259377323715632728573145444117676200197965533814287619148566636993448985613167
30195956933144181457716970689297269742173162426908289983779501164116695096053678
91007422675363443658034720511968591074370394031117377266157313625740076978761478
86258594553557393376919388439633643700224
which, divided by 4096 leaves a remainder of 1024. In fact the UBASIC interpreter which gives this, has directly a powermod function:
?modpow(2,1990,1990)
1024
OK

Posted by Charlie on 2003-03-02 09:21:08