Problem 4. Determine the smallest consecutive natural numbers $a, b, c, d$ knowing that $a$ is divisible by $8$, $b$ by $7$, $c$ by $6$, and $d$ by $5$.
Solution. Let $a, b=a+1, c=a+2, d=a+3$ be the four consecutive numbers. We obtain:

$$
\begin{aligned}
& 8 \mid a \Longrightarrow a=8 m \\
& 7|a+1=8 m+1=7 m+(m+1) \Longrightarrow 7| m+1 \\
& 6|a+2=8 m+2=6 m+2(m+1) \Longrightarrow 6| 2(m+1) \Longrightarrow 3 \mid m+1 \\
& 5|a+3=8 m+3=5 m+3(m+1) \Longrightarrow 5| 3(m+1) \Longrightarrow 5 \mid m+1
\end{aligned}
$$

From (4), (5) and (6) it follows that $3 \cdot 5 \cdot 7=105 \mid m+1 \Longrightarrow m+1=105 k, k \in N^{*}$. Since $a$ must be the smallest, we take $k=1$ so $m=104$.

We obtain $a=832, b=833, c=834, d=835$.

## GRADING SCALE

- 5th grade -