"Concantenate" is a long word

For a number to be divisible by 3, the sum of its digits must be a multiple of 3.

Concantenation multiplies our 1729-digit number by $\displaystyle \sum_{k=0}^{1728} 10^{1729k}$ , which is not divisible by 3 since 1729 is not; so, our 1729-digit number must have a factor of 3 instead.

Let $n_{j}$ be the number of elements of $\lbrace a_{i} \rbrace$ congruent to $j$ mod 3. This means that $n_{2} = 490 \equiv 1 \pmod{3}$ .

$a_{i}$ can either be 0, 1 or 2 mod 3. However, $a_{i}^{2}$ can only be 0 or 1 mod 3 (easy check with $0^{2}$ , $1^{2}$ , and $2^{2}$ suffices). Thus,

$49000 = \sum_{n=1}^{1729} a_{i}^{2} \equiv 0n_{0}+1n_{1}+1n_{2} \equiv n_{1}+1 \pmod{3}$

which implies $n_{1} \equiv 0 \pmod{3}$ .

The sum of digits of our 1729-digit number is then

$\sum_{n=1}^{1729} a_{i} \equiv 0n_{0}+1n_{1}+2n_{2} \equiv 2 \pmod{3}$

As such, no $N$ divisible by 3 exists.

From the given hint 0<ai<9,we can say that no number of N is divisible by 3;hence answer is 0