how-to-prove-that-the-number-is-divisible-by-3-then-the-number-of-numbers-is-a-multiple-of-3-

Question Number 81647 by john santu last updated on 14/Feb/20

$$\mathrm{how}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{number}\: \\ $$$$\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3},\:\mathrm{then}\:\mathrm{the}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{numbers}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3} \\ $$

Commented by mr W last updated on 14/Feb/20

did you want to say: how to prove that a number is divisible by 3 when the sum of its digits is a multiple of 3. or how to prove that, if a number is divisible by 3, then the sum of its digits is a multiple of 3.

$${did}\:{you}\:{want}\:{to}\:{say}: \\ $$$${how}\:{to}\:{prove}\:{that}\:{a}\:{number}\:{is}\:{divisible} \\ $$$${by}\:\mathrm{3}\:{when}\:{the}\:{sum}\:{of}\:{its}\:{digits}\:{is}\:{a} \\ $$$${multiple}\:{of}\:\mathrm{3}. \\ $$$${or} \\ $$$${how}\:{to}\:{prove}\:{that},\:{if}\:{a}\:{number}\:{is}\:{divisible} \\ $$$${by}\:\mathrm{3},\:{then}\:{the}\:{sum}\:{of}\:{its}\:{digits}\:{is}\:{a} \\ $$$${multiple}\:{of}\:\mathrm{3}. \\ $$

Commented by Kunal12588 last updated on 14/Feb/20

100a+10b+c=3q for a 3 digit no. ⇒3q=99a+9b+(a+b+c) ⇒q=33a+3b+((a+b+c)/3) q ∈ N ⇒ a+b+c ∣ 3 for a ′n′ digit no. 3m=Σ_(i=0) ^(n−1) 10^i a_i ; a_i is (i+1)^(th) digit ⇒m=3k + (1/3)Σ_(i=0) ^(n−1) a_i ⇒Σ_(i=0) ^(n−1) a_i is divisible by 3. please help me to find k

$$\mathrm{100}{a}+\mathrm{10}{b}+{c}=\mathrm{3}{q}\:\:\:{for}\:{a}\:\mathrm{3}\:{digit}\:{no}. \\ $$$$\Rightarrow\mathrm{3}{q}=\mathrm{99}{a}+\mathrm{9}{b}+\left({a}+{b}+{c}\right) \\ $$$$\Rightarrow{q}=\mathrm{33}{a}+\mathrm{3}{b}+\frac{{a}+{b}+{c}}{\mathrm{3}} \\ $$$${q}\:\in\:{N}\:\Rightarrow\:{a}+{b}+{c}\:\mid\:\mathrm{3} \\ $$$${for}\:{a}\:'{n}'\:{digit}\:{no}. \\ $$$$\mathrm{3}{m}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{10}^{{i}} {a}_{{i}} \:;\:{a}_{{i}} \:{is}\:\left({i}+\mathrm{1}\right)^{{th}} \:{digit} \\ $$$$\Rightarrow{m}=\mathrm{3}{k}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{a}_{{i}} \: \\ $$$$\Rightarrow\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{a}_{{i}} \:\:{is}\:{divisible}\:{by}\:\mathrm{3}. \\ $$$${please}\:{help}\:{me}\:{to}\:{find}\:{k} \\ $$

Commented by Prithwish Sen 1 last updated on 14/Feb/20

any number can be represented by N=10^n x_n +10^(n−1) x_(n−1) +........+10^2 x_2 +10x_1 +x_0 now x_n +x_(n−1) +...........+x_1 +x_0 = 3k ∴ N= (999...ntimes+1)x_n +......+(99+1)x_2 +(9+1)x_1 +x_0 = (999...n times .x_n + ....+99x_2 +9x_1 )+(x_n +x_(n−1) +.....+x_2 +x_1 +x_0 ) =3m+3k i.e N is also divisible by 3 (proved)

$${any}\:{number}\:{can}\:{be}\:{represented}\:{by}\: \\ $$$$\boldsymbol{{N}}=\mathrm{10}^{{n}} {x}_{{n}} +\mathrm{10}^{{n}−\mathrm{1}} {x}_{{n}−\mathrm{1}} +……..+\mathrm{10}^{\mathrm{2}} {x}_{\mathrm{2}} \:+\mathrm{10}{x}_{\mathrm{1}} +{x}_{\mathrm{0}} \\ $$$${now} \\ $$$${x}_{{n}} +{x}_{{n}−\mathrm{1}} +………..+{x}_{\mathrm{1}} +{x}_{\mathrm{0}} =\:\mathrm{3}\boldsymbol{{k}} \\ $$$$\therefore\:\boldsymbol{{N}}=\:\left(\mathrm{999}…{ntimes}+\mathrm{1}\right){x}_{{n}} +……+\left(\mathrm{99}+\mathrm{1}\right){x}_{\mathrm{2}} +\left(\mathrm{9}+\mathrm{1}\right){x}_{\mathrm{1}} +{x}_{\mathrm{0}} \\ $$$$=\:\left(\mathrm{999}…{n}\:{times}\:.{x}_{{n}} +\:….+\mathrm{99}{x}_{\mathrm{2}} +\mathrm{9}{x}_{\mathrm{1}} \right)+\left({x}_{{n}} +{x}_{{n}−\mathrm{1}} +…..+{x}_{\mathrm{2}} +{x}_{\mathrm{1}} +{x}_{\mathrm{0}} \right) \\ $$$$=\mathrm{3}\boldsymbol{{m}}+\mathrm{3}\boldsymbol{{k}} \\ $$$$\boldsymbol{{i}}.\boldsymbol{{e}}\:\boldsymbol{{N}}\:\boldsymbol{{is}}\:\boldsymbol{{also}}\:\boldsymbol{{divisible}}\:\boldsymbol{{by}}\:\mathrm{3}\:\left(\boldsymbol{{proved}}\right) \\ $$$$ \\ $$

Commented by mr W last updated on 14/Feb/20

3m=Σ_(i=0) ^(n−1) 10^i a_i 3m=Σ_(i=0) ^(n−1) (1+9)^i a_i 3m=Σ_(i=0) ^(n−1) [Σ_(k=0) ^i C_k ^i 9^k ]a_i 3m=Σ_(i=0) ^(n−1) [1+Σ_(k=1) ^i C_k ^i 9^k ]a_i 3m=Σ_(i=0) ^(n−1) a_i +Σ_(i=1) ^(n−1) (Σ_(k=1) ^i C_k ^i 3^(2k) )a_i 3m=Σ_(i=0) ^(n−1) a_i +3h ⇒Σ_(i=0) ^(n−1) a_i =3(m−h)=multiple of 3

$$\mathrm{3}{m}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\mathrm{10}^{{i}} {a}_{{i}} \\ $$$$\mathrm{3}{m}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\mathrm{1}+\mathrm{9}\right)^{{i}} {a}_{{i}} \\ $$$$\mathrm{3}{m}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left[\underset{{k}=\mathrm{0}} {\overset{{i}} {\sum}}{C}_{{k}} ^{{i}} \mathrm{9}^{{k}} \right]{a}_{{i}} \\ $$$$\mathrm{3}{m}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left[\mathrm{1}+\underset{{k}=\mathrm{1}} {\overset{{i}} {\sum}}{C}_{{k}} ^{{i}} \mathrm{9}^{{k}} \right]{a}_{{i}} \\ $$$$\mathrm{3}{m}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{a}_{{i}} +\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\underset{{k}=\mathrm{1}} {\overset{{i}} {\sum}}{C}_{{k}} ^{{i}} \mathrm{3}^{\mathrm{2}{k}} \right){a}_{{i}} \\ $$$$\mathrm{3}{m}=\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{a}_{{i}} +\mathrm{3}{h} \\ $$$$\Rightarrow\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{a}_{{i}} =\mathrm{3}\left({m}−{h}\right)={multiple}\:{of}\:\mathrm{3} \\ $$

Commented by Prithwish Sen 1 last updated on 14/Feb/20