Question Number 129087 by Adel last updated on 12/Jan/21 | ||
Answered by MJS_new last updated on 12/Jan/21 | ||
$$\mathrm{2}^{\mathrm{10}} \mathrm{10}!\left(\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{7}×\mathrm{9}×\mathrm{11}×\mathrm{13}×\mathrm{15}×\mathrm{17}×\mathrm{19}\right)= \\ $$$$=\mathrm{10}!×\mathrm{11}×\left(\mathrm{3}×\mathrm{2}^{\mathrm{2}} \right)×\mathrm{13}×\left(\mathrm{7}×\mathrm{2}\right)×\mathrm{15}×\left(\mathrm{2}^{\mathrm{4}} \right)×\mathrm{17}×\left(\mathrm{9}×\mathrm{2}\right)×\mathrm{19}×\left(\mathrm{5}×\mathrm{2}^{\mathrm{2}} \right)= \\ $$$$=\mathrm{20}! \\ $$ | ||