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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
howtoprovethatthenumberisdivisibleby3,thenthenumberofnumbersisamultipleof3
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.
didyouwanttosay:howtoprovethatanumberisdivisibleby3whenthesumofitsdigitsisamultipleof3.orhowtoprovethat,ifanumberisdivisibleby3,thenthesumofitsdigitsisamultipleof3.
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
100a+10b+c=3qfora3digitno.3q=99a+9b+(a+b+c)q=33a+3b+a+b+c3qNa+b+c3forandigitno.3m=n1i=010iai;aiis(i+1)thdigitm=3k+13n1i=0ain1i=0aiisdivisibleby3.pleasehelpmetofindk
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)
anynumbercanberepresentedbyN=10nxn+10n1xn1+..+102x2+10x1+x0nowxn+xn1+..+x1+x0=3kN=(999ntimes+1)xn++(99+1)x2+(9+1)x1+x0=(999ntimes.xn+.+99x2+9x1)+(xn+xn1+..+x2+x1+x0)=3m+3ki.eNisalsodivisibleby3(proved)
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
3m=n1i=010iai3m=n1i=0(1+9)iai3m=n1i=0[ik=0Cki9k]ai3m=n1i=0[1+ik=1Cki9k]ai3m=n1i=0ai+n1i=1(ik=1Cki32k)ai3m=n1i=0ai+3hn1i=0ai=3(mh)=multipleof3
Commented by Prithwish Sen 1 last updated on 14/Feb/20
How are you sir ?
Howareyousir?
Commented by mr W last updated on 14/Feb/20
every thing is ok with me, thanks!   you too sir?
everythingisokwithme,thanks!youtoosir?
Commented by john santu last updated on 14/Feb/20
yes.
yes.
Commented by Prithwish Sen 1 last updated on 15/Feb/20
fine sir.
finesir.

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