Question Number 178213 by Acem last updated on 14/Oct/22
$${Let}\:{S}=\:\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:...,\:\mathrm{28},\:\mathrm{29},\:\mathrm{30}\right\} \\ $$$$\:{How}\:{many}\:{subgroups}\:{of}\:\mathrm{3}\:{elements}\:{are}\:{in}\:{the}\:{S} \\ $$$$\:\:{so}\:{that}\:{their}\:{sum}\:{is}\:{a}\:{multiple}\:{of}\:\mathrm{3} \\ $$$$ \\ $$$$\:{The}\:{answer}\:{is}\:\mathrm{1}\:\mathrm{360}\:{subgroup}\mathrm{s} \\ $$$$ \\ $$
Commented by Acem last updated on 14/Oct/22
$$\boldsymbol{{Hint}}:\:{n}\:{mod}\:\mathrm{3}\:=\:\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{2}\right\}\:\forall{n}\in\:\mathbb{N} \\ $$
Answered by nikif99 last updated on 14/Oct/22
$${Let}\:{S}_{\mathrm{0}} =\left\{\mathrm{3},\:\mathrm{6},\:\mathrm{9},\:...,\:\mathrm{30}\right\}\:{n}=\mathrm{10} \\ $$$${S}_{\mathrm{1}} =\left\{\mathrm{1},\:\mathrm{4},\:\mathrm{7},\:...,\:\mathrm{28}\right\}\:{n}=\mathrm{10} \\ $$$${S}_{\mathrm{2}} =\left\{\mathrm{2},\:\mathrm{5},\:\mathrm{8},\:...,\:\mathrm{29}\right\}\:{n}=\mathrm{10} \\ $$$${Sum}\:{divisible}\:{by}\:\mathrm{3}: \\ $$$$\left.{a}\right)\:\mathrm{1}\:{element}\:{from}\:{each}\:{S}_{\mathrm{0}} ,\:{S}_{\mathrm{1}} ,\:{S}_{\mathrm{2}} :\:\mathrm{10}^{\mathrm{3}} \\ $$$$\left.{b}\right)\:\mathrm{3}\:{elements}\:{from}\:{S}_{\mathrm{0}} :\:{C}_{\mathrm{3}} ^{\mathrm{10}} =\mathrm{120} \\ $$$$\left.{c}\right)\:\mathrm{3}\:{elements}\:{from}\:{S}_{\mathrm{1}} :\:{C}_{\mathrm{3}} ^{\mathrm{10}} =\mathrm{120} \\ $$$$\left.{d}\right)\:\mathrm{3}\:{elements}\:{from}\:{S}_{\mathrm{2}} :\:{C}_{\mathrm{3}} ^{\mathrm{10}} =\mathrm{120} \\ $$$${Total}\:\mathrm{1360} \\ $$
Commented by Acem last updated on 14/Oct/22
$${A}\:{bouquet}\:{of}\:{flowers}! \\ $$
Commented by nikif99 last updated on 14/Oct/22
$${Thank}\:{you}. \\ $$
Commented by Tawa11 last updated on 15/Oct/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$