The biggest value of k

[rakeen]

if $N$ $=$ 19202123.....909192

and N | $3^k$ then what is the biggest value of $k$ in +ve integer?

I found that the sum of the digits are 684 , which is divisible by 3 & 9.

[tushar7]

so $ k=2$

[AntiviruShahriar]

sum of digits amar astese $670$......$k=0$.....
dekho sum ta:
$1$ ashe $9$ bar-v19,21,31,41,51,61,71,81,91....2$ ashe $17v bar $20=29,32,42,52,62,72,82,92....$
$3$ theke $8$ porjonto egula ashe $16$ bar......$9$ ashe 11 bar-$19,29,39,49,59,79,89,90,91,92...$
so sum$=(1*9)+(2*17)+16*(3+4+5+6+7+8)+9*11=670$

[rakeen]

@Anti:is should not be. coz if it is 670 then it won't be divisible by 3. And the question says it must be divisible by $3^k$. If it is 670, then k must be 0.

[rakeen]

@Tushar it's not necessary. coz it might be divisible by 27 too. i.e. 1512 is divisible by 9. coz 1+5+1+2=9. But it is also bivisible by 27 also. 27*56=1512

[Avik Roy]

You precisely don't need to calculate the sum of the digits. If $0 \le k \le 2$ you can find it by simply evaluating the sum $19+20+...92$ which is $4278$ this is divisible by $3$ but not $9$. That makes the result being $1$

[rakeen]

I didn't understand.

[AntiviruShahriar]

You precisely don't need to calculate the sum of the digits. If $0 \le k \le 2$ you can find it by simply evaluating the sum $19+20+...92$ which is $4278$ this is divisible by $3$ but not $9$. That makes the result being $1$

by simply evaluating the sum $19+20+...92$ which is $4107$ this is not divisible by both $3$ and $9$...That makes the result being $0$

[Avik Roy]

I misevaluated the sum...
however, $4107=3.1369$ which implies that it is divisible by $3$. The result is $1$

This question is from BdMO national of 2007, I was an attendee there and I know the result right

[rakeen]

hey, you didn't tell me why are we using 19+20+21+22+....
if we add all the digits the result is divisible by 9 also.