1.a) Find 11 consecutive natural numbers whose sum is 99.

b) A four-digit natural number has its first two digits identical, and the units digit is 5. This number is divided by a two-digit number, and the remainder is 98. Find the dividend, divisor, and quotient.

Prof. Ana Marcela Popa
## Solution:

a) $\quad(a+1)+(a+2)+(a+3)+\ldots+(a+11)=99 \Leftrightarrow 11 a+66=99 \Leftrightarrow a=3$. The numbers are $4,5,6, \ldots, 14$

b) We denote $\overline{a a b 5}, \mathrm{a} \neq 0$ as the natural number considered.

$\overline{a a b 5}=\overline{x y} \cdot \mathrm{q}+98,0 \leq 98<\overline{x y} \Rightarrow \overline{x y}=99$.

$\overline{a a b 5}=99 \cdot \mathrm{q}+98 . \Rightarrow$ the last digit of $99 \cdot \mathrm{q}+98$ is 5 and thus the last digit of q is 3.

$1105 \leq \overline{a a b 5} \leq 9995 \Rightarrow 1105 \leq 99 \cdot \mathrm{q}+98 \leq 9995 \Rightarrow 11 \leq \mathrm{q} \leq 99$ and, since the last digit of q is 3, $\Rightarrow \mathrm{q} \in\{13,23,33, \ldots, 93\}$.

The dividend is 3365, the divisor is 99 and the quotient is 33.

## Scoring Guide

| a) | $(\mathrm{a}+1)+(\mathrm{a}+2)+(\mathrm{a}+3)+\ldots+(\mathrm{a}+11)=99$ | $2 p$ |
| :--- | :--- | :--- |
| The numbers are $4,5,6, \ldots, 14$ | $1 p$ |  |
| b) | $\overline{a a b 5}=\mathrm{x} \cdot \mathrm{q}+98,0 \leq 98<\mathrm{x}$ | $1 p$ |
| the last digit of $99 \cdot \mathrm{q}+98$ is 5 and $\mathrm{U}(\mathrm{q})=3$ | $1 p$ |  |
| Verification for $\mathrm{q} \in\{13,23,33, \ldots, 93\}$ | $1 p$ |  |
| The dividend is 3365, the divisor is 99 and the quotient is 33 | $1 p$ |  |