Jutaro wrote: i) Show that there exists a number $N$ such that its first 2003 multiples $N,\ 2N,\cdots,\ 2003N$ are all tico. $N=100010001\cdot\cdot\cdot 0001$ works! Where number of $1's$ is equal to $2003$ and between two $1's$ have three $0's$.

ii) The answer of this part is negative. Let $S(n)$ denote the sum of digits of $n$. Let $N$ a natural number, it`s well known that $N$ has a multiple of the form $999\ldots 000$, let \[M=\underbrace{99\ldots 9}_{m}\underbrace{00\ldots 0 }_{k}\] this multiple (of $N$). Note that $S(M)=9m$. The number \[P=\underbrace{99\ldots 9}_{m+1}M=\underbrace{99\ldots 9}_{m-1}89\underbrace{00\ldots 0}_{m-1}1\underbrace{00\ldots 0}_{k}\] is also a multiple of $N$ and $S(P)=9(m+1)$. Finally, since $S(P)-S(M)=9$ and both, $P$ and $M$, are multiple of $N$ someone of them is not tico. $Tipe$

(1)we consider $N=10000100001......100001$ ($2003$ 1s in total)

(2) if $(n,10)=1$ there exists $k$,$10^k\equiv 1(mod n)$ so $10^{nk}+10^{(n-1)k}+......+10^k$ is a multiple of $n$,its digit sum is $n$ if n is not a multiple of $2003$,proved if $2003|n$,trivially $n>10$ hence $10^{nk}+10^{(n-1)k}+......+10^k$ is a multiple of $n$ its digit sum is $n-9 \equiv -9(mod 2003)$ proved if $(n,10)>1$,proved analagously QED

N.T.TUAN wrote: $N=100010001\cdot\cdot\cdot 0001$ works! Where number of $1's$ is equal to $2003$ and between two $1's$ have three $0's$. Where does this idea come from?