# "Concantenate" is a long word

Number Theory Level 5

I take the $$1729$$-digit number $$\overline{a_{1}a_{2} \ldots a_{1729}}$$ and concantenate it with itself $$1729$$ times to make a new $$1729^{2}$$-digit number

$N = \underbrace{\overline{a_{1}a_{2} \ldots a_{1729}a_{1}a_{2} \ldots a_{1729} \ldots \ldots a_{1}a_{2} \ldots a_{1729}}}_{1729 \ \text{times}}$

If the sequence $$\lbrace a_{i} \rbrace$$ satisfies that $$\displaystyle \sum_{i=1}^{1729} a_{i}^{2} = 49000$$ and there are exactly $$490$$ solutions $$i$$ to the congruence $$a_{i} \equiv 2 \pmod{3}$$, find the number of $$N$$ divisible by $$3$$.

Note: $$0 \leq a_{i} \leq 9$$.

×