dwie ostatnie cyfry
: 2 cze 2015, o 22:53
wyznaczyć 2 ostatnie cyfry liczby \(\displaystyle{ 51 ^{1000}-99 ^{77}}\)
\(\displaystyle{ 51\equiv 51\pmod{100}}\)/ \(\displaystyle{ ^{2}}\)
\(\displaystyle{ 51 ^{2} \equiv 2601\pmod{100}}\)
\(\displaystyle{ 51 ^{2} \equiv 1\pmod{100}}\)
\(\displaystyle{ 51 ^{1000} \equiv 1 ^{1000} \pmod{100}}\)
\(\displaystyle{ 51 ^{1000} \equiv 1\pmod{100}}\)
\(\displaystyle{ 99 \equiv -1\pmod{100}}\)/\(\displaystyle{ ^{77}}\)
\(\displaystyle{ 99 ^{77} \equiv -1 ^{77} \pmod{100}}\)
\(\displaystyle{ 99 ^{77} \equiv -1\pmod{100}}\)
Dobrze to jest policzone?
\(\displaystyle{ 51\equiv 51\pmod{100}}\)/ \(\displaystyle{ ^{2}}\)
\(\displaystyle{ 51 ^{2} \equiv 2601\pmod{100}}\)
\(\displaystyle{ 51 ^{2} \equiv 1\pmod{100}}\)
\(\displaystyle{ 51 ^{1000} \equiv 1 ^{1000} \pmod{100}}\)
\(\displaystyle{ 51 ^{1000} \equiv 1\pmod{100}}\)
\(\displaystyle{ 99 \equiv -1\pmod{100}}\)/\(\displaystyle{ ^{77}}\)
\(\displaystyle{ 99 ^{77} \equiv -1 ^{77} \pmod{100}}\)
\(\displaystyle{ 99 ^{77} \equiv -1\pmod{100}}\)
Dobrze to jest policzone?