## SUBJECT 1

Let $n$ be a natural number. If we place the digit 7 in front of $n$, we get a number five times larger than when we place the digit 7 at the end of $n$. Find the smallest value of $n$.

GM 11/2013(SUPPLEMENT)
SUBIECTUL 1

Let $n$ be a natural number. If we place the digit 7 in front of $n$, we get a number five times larger than when we place the digit 7 at the end of $n$. Find the smallest value of $n$.

| From the relation $\overline{7 \mathrm{n}}=5 \cdot \overline{\mathrm{n} 7} \Rightarrow u(\mathrm{n})=5$ | $2 \mathrm{p}$ |
| :--- | :--- |
| We obtain the relation $\overline{7 \mathrm{a} 5}=5 \cdot \overline{\mathrm{a} 57} \Rightarrow u(\mathrm{a})=8$ | $1 \mathrm{p}$ |
| We have: $\overline{7 \mathrm{~b} 85}=5 \cdot \overline{\mathrm{b} 857} \Rightarrow u(\mathrm{~b})=2$ | $1 \mathrm{p}$ |
| Then: $\overline{7 \mathrm{~c} 285}=5 \cdot \overline{\mathrm{c} 2857} \Rightarrow \mathrm{u}(\mathrm{c})=4$ | $1 \mathrm{p}$ |
| And then: $\overline{7 \mathrm{~d} 4285}=5 \cdot \overline{\mathrm{d} 42857}$, obtaining $d=1$ | $1 \mathrm{p}$ |
| We obtain the smallest value of $n=14285$. | $1 \mathrm{p}$ |

OBS. If the analysis is correct for $n$ of one digit, $n$ of two digits, etc., 1 p is awarded for each case and a maximum after reaching the final equation $\overline{7 \mathrm{abcde}}=5 \cdot \overline{\mathrm{abcde7}}$ and solving it to get $\overline{\mathrm{abcde}}=14285$.