38.3. Find a four-digit number that is 4 times less than the number written with the same digits but in reverse order.

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\text { (8-10 grades) }
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38.3. It is required to find the number $\overline{a b c e}$ such that $4 \cdot \overline{a b c e}=\overline{e c b a}$. Since the product $4 \cdot \overline{a b c e}$ is even, the number $\overline{e c b a}$ is also even, i.e., the digit $a$ is even. Then it is easy to see that $a=2$ (since for $a>2$ the product $4 \cdot \overline{a b c e}$ would be a five-digit number). Moreover, since the product $4 \cdot e$ must end in $a$, i.e., in 2, then $e=8$. Thus, $4 \cdot \overline{2 b c 8}=\overline{8 c b 2}$. From this it follows that $4 \cdot \overline{b c}+3=\overline{c b}$, i.e., $40 b+4 c+3=10 c+b$ or $13 b+1=2 c$. The obtained equation for the digits $b$ and $c$ has a unique solution: $b=1, c=7$. Answer: $\overline{a b c e}=2178$.