Ответ:

[tex]\large \boldsymbol {}\mathrm {HOD} (2^{2^{123} -1} - 1 ~ , ~ 2^{2^{321}-1}-1 ) = 127[/tex]

Объяснение:

Воспользуемся теоремой

[tex]\large \boldsymbol {}\mathrm {HOD} (a^n - 1 ~ , ~ a^m - 1) = a^{\mathrm{HOD}(n , m ) }-1[/tex]

Следовательно

[tex]\large \boldsymbol {}\mathrm {HOD} (2^{2^{123} -1} - 1 ~ , ~ 2^{2^{321}-1}-1 ) = 2^{\mathrm{HOD}(2^{123} -1~ , ~2^{321}-1 ) }-1[/tex]

По той же теореме

[tex]\large \boldsymbol{} \mathrm{HOD} (2^{123} -1~ , ~2^{321}-1 ) = \mathrm{HOD} (2^{41\cdot 3} -1~ , ~2^{107\cdot 3}-1 ) =\\\\= 2^3 - 1 = 7[/tex]

[tex]\large \boldsymbol {}\mathrm {HOD} (2^{2^{123} -1} - 1 ~ , ~ 2^{2^{321}-1}-1 ) = 2^{\mathrm{HOD}(2^{123} -1~ , ~2^{321}-1 ) }-1 = 2^7 -1 = 127[/tex]