400**. Can the digits of a prime number not exceeding $10^{4}$ form an arithmetic or geometric progression? If they can, find these numbers.
Solution. 1) Suppose the digits of the number form an arithmetic progression.

a) For a three-digit number, we will have the digits: $x, x+d$, $x+2d$, where $d$ is the difference of the progression. Their sum $x+(x+d)+(x+2d)$ will be divisible by 3, i.e., the number formed by these digits is divisible by 3 and cannot be prime, which contradicts the problem's condition.

b) For a four-digit number, we will have the digits: $x, x+d$, $x+2d, x+3d$. The sum of the numbers represented by these digits will be:

$$
x+x+d+x+2d+x+3d=2(2x+3d)
$$

$x$ cannot be equal to 3, because then we would get a number divisible by 3, i.e., not prime. Further, since $x+3d$ is a digit of the number, then $x+3d \leqslant 9$ and $d$ can be 2 or 1.

Let $d=2$. Then $x \leqslant 3 ; x=3$ cannot be (point b); $x=2$ also cannot be, because then we would get that the given number is even; $x=1$ can be, then we get the numbers: 1357 and 7531, but both these numbers are composite.

Let $d=1$. We get the numbers: 1234, 4321, 2345, 5432, 3456, 6543, 4567.

Numbers: 1234, 5432, 3456 - are even; 4321 is divisible by 29; 2345 is divisible by 5; 6543 is divisible by 3. There is one prime number 4567.

2) Suppose the digits of the number form a geometric progression, i.e., $a, aq, aq^2, aq^3$.

a) For a three-digit number $aq^2 \leqslant 9$, where $q$ is the common ratio of the progression forming the digits of the number. Thus, $q \leqslant 3$, and if the hundreds digit is 1, then the only possibility is the number 139.

If the hundreds digit is not 1, then $q$ cannot be an integer, because in this case the given number would be divisible by the number represented by the first digit and would not be a prime number.

Let $q=\frac{p}{n}$, and $p$ and $n$ cannot be greater than 3, and $q$ can have the values: $\frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{3}{2}$. We get the numbers: 469, 964, 421, 842, 931. Of these, only 421 is a prime number.

b) For a four-digit number, we will have: $aq^3 \leqslant 9$ The only value for $q$ is $q=2$ and $a=1$. We get the number 8421, which is divisible by 3.

Thus, there can only be three numbers that satisfy the problem's condition: 4567 (arithmetic progression), 139 and 421 (geometric progression).

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## § 26. Harmonic Proportion and Harmonic Progression*