BdMO National Higher Secondary 2007/6

[BdMO]

Problem 6:
Writing down all the integers from $19$ to $92$ we make a large integer $N$.\[N=192021\cdots 909192\]If $N$ is divisible by $3^k$ then what is the maximum value of $k$?

[bristy1588]

Can anyone tell me how to proceed in these types of Problems?

[*Mahi*]

Problem 6:
Writing down all the integers from $19$ to $92$ we make a large integer $N$.\[N=192021\cdots 909192\]If $N$ is divisible by $3^k$ then what is the maximum value of $k$?

Here it is to be used that if the sum of digits of $N$ is divisible by $3^k$, then so is $N$.

[sourav das]

Here it is to be used that if the sum of digits of $N$ is divisible by $3^k$, then so is $N$.

Mahi, is it true for $k>2$ ?Of course, it can be used for checking that is $k<2$ or not. Because when the sum of all digit of "N" can divisible by 9 then $k$ can be greater or equal $2$.

[nafistiham]

$3^3=27$, but, $27$ does not divide $2+7=9$

[*Mahi*]

Here it is to be used that if the sum of digits of $N$ is divisible by $3^k$, then so is $N$.

Mahi, is it true for $k>2$ ?Of course, it can be used for checking that is $k<2$ or not. Because when the sum of all digit of "N" can divisible by 9 then $k$ can be greater or equal $2$.

This theorem can be improvised, like $1000 \equiv 1 mod 27$

[sourav das]

Actually, i just want to say we just can check whether N is divisible by 9 or not for interesting digit pattern and if satisfy, we can think other ways but not in digit way. It is easy to check that $N$ is not divisible by 9. As 19+20+....+91=37*111, which is only divisible by 3, not 9.

[*Mahi*]

Thanks for opening your eyes.