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

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

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

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

How are you sir ?

$${How}\:{are}\:{you}\:{sir}\:? \\ $$

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

every thing is ok with me, thanks!   you too sir?

$${every}\:{thing}\:{is}\:{ok}\:{with}\:{me},\:{thanks}!\: \\ $$$${you}\:{too}\:{sir}? \\ $$

Commented by john santu last updated on 14/Feb/20

yes.

$$\mathrm{yes}.\: \\ $$

Commented by Prithwish Sen 1 last updated on 15/Feb/20

fine sir.

$${fine}\:{sir}. \\ $$

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