4. In the addition problem, different letters represent different digits. In the sum (which is written with the word NINA), all digits are odd. What is the smallest possible value of the sum?

| $R$ | $E$ | $P$ | $O$ |
| ---: | ---: | ---: | ---: |
| $+\quad R$ | $I$ | $B$ | $A$ |
| $N$ | $I$ | $N$ | $A$ |

MATHEMATICAL COMPETITION

April 7, 2001

## PROBLEMS FOR THE THIRD YEAR
II/4. Since $O+A=A$, it follows that $O=0$. Therefore, $E=9$ and $P+B>10$ (since $P+B \neq 10$, because $N \neq 0$). Then $N=2R+1$. $N$ can be at least 3. In this case, $R=1$ and $P+B=13$, which means that $P$ and $B$ are either 7 and 6 or 8 and 5 (in any order). However, $I$ and $A$ should be different odd digits, but we do not have enough of them available. Therefore, we choose $N=5$. Then $R=2$ and $P+B=15$, which means that the digits $P$ and $B$ are 7 and 8 (in any order). Since we are looking for the smallest possible sum, we choose $I=1$ and $A=3$, and we have the solution 5153.

Findings: $O=0, E=9, N$ odd: 1 point each. Rejected option $N=3$: 2 points. Determined the smallest sum 5153 for the option $N=5$: 2 points (only 1 point if not all unknown digits in the equation are specified). If the contestant only guesses the solution 5153 but does not prove its minimality: 2 points.