Let-S-1-2-3-4-28-29-30-How-many-subgroups-of-3-elements-are-in-the-S-so-that-their-sum-is-a-multiple-of-3-The-answer-is-1-360-subgroups-

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_0 ={3, 6, 9, ..., 30} n=10 S_1 ={1, 4, 7, ..., 28} n=10 S_2 ={2, 5, 8, ..., 29} n=10 Sum divisible by 3: a) 1 element from each S_0 , S_1 , S_2 : 10^3 b) 3 elements from S_0 : C_3 ^(10) =120 c) 3 elements from S_1 : C_3 ^(10) =120 d) 3 elements from S_2 : C_3 ^(10) =120 Total 1360

$${Let}\:{S}_{\mathrm{0}} =\left\{\mathrm{3},\:\mathrm{6},\:\mathrm{9},\:…,\:\mathrm{30}\right\}\:{n}=\mathrm{10} \\ $$$${S}_{\mathrm{1}} =\left\{\mathrm{1},\:\mathrm{4},\:\mathrm{7},\:…,\:\mathrm{28}\right\}\:{n}=\mathrm{10} \\ $$$${S}_{\mathrm{2}} =\left\{\mathrm{2},\:\mathrm{5},\:\mathrm{8},\:…,\:\mathrm{29}\right\}\:{n}=\mathrm{10} \\ $$$${Sum}\:{divisible}\:{by}\:\mathrm{3}: \\ $$$$\left.{a}\right)\:\mathrm{1}\:{element}\:{from}\:{each}\:{S}_{\mathrm{0}} ,\:{S}_{\mathrm{1}} ,\:{S}_{\mathrm{2}} :\:\mathrm{10}^{\mathrm{3}} \\ $$$$\left.{b}\right)\:\mathrm{3}\:{elements}\:{from}\:{S}_{\mathrm{0}} :\:{C}_{\mathrm{3}} ^{\mathrm{10}} =\mathrm{120} \\ $$$$\left.{c}\right)\:\mathrm{3}\:{elements}\:{from}\:{S}_{\mathrm{1}} :\:{C}_{\mathrm{3}} ^{\mathrm{10}} =\mathrm{120} \\ $$$$\left.{d}\right)\:\mathrm{3}\:{elements}\:{from}\:{S}_{\mathrm{2}} :\:{C}_{\mathrm{3}} ^{\mathrm{10}} =\mathrm{120} \\ $$$${Total}\:\mathrm{1360} \\ $$

Commented by Acem last updated on 14/Oct/22