## SUBJECT IV

We have $n=\overline{11 \ldots 1}+\overline{22 \ldots 2}+\ldots+\overline{88 \ldots 8}+\overline{99 \ldots 9}=(1+2+\ldots+8+9) \cdot \overline{11 \ldots 1}$ $\qquad$ 1 point

$1+2+\ldots+8+9=45$ $\qquad$
2 points .

We obtain $n=45 \cdot \overline{11 \ldots 1}$, $\qquad$
which leads to $n=\overline{499 \ldots .95}$, a number with 2016 digits $\qquad$ 1 point .

Finalization, $n$ contains 2014 nines 1 point .

Grading: $7 p$
## SUBJECT IV

Let the natural number $n=\overline{11 \ldots 1}+\overline{22 \ldots 2}+\ldots+\overline{88 \ldots 8}+\overline{99 \ldots 9}$, where each number of the form $\overline{a a \ldots a}$ contains 2015 digits of $a$. Determine how many digits of 9 the number $n$ contains.

## Note:

All subjects are mandatory;

Each subject is graded out of 7 points;

No points are awarded by default;

The actual working time is 2 hours from the moment the subject is received.

\section*{MATHEMATICS OLYMPIAD LOCAL STAGE 14.02.2015

GRADE 5  Solutions and grading criteria}

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CLASA a V-a  Soluții şi barem de corectare} 

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