The sum of two numbers is 173,717. The prime factors of the four-digit difference between the two numbers do not include any single-digit numbers. One of the numbers is divisible by 1558. Which are these two numbers?
I. solution: One of the numbers is obviously smaller than half the sum by the same amount that the other is larger than half the sum. Since the difference between the two numbers is less than 10000, the sought numbers, if we write $m$ instead of $\frac{173177}{2}=86858.5$, are between $m-5000$ and $m+5000$, that is, between 81859 and 91859.

Let's find the first number divisible by 1558 in the sequence of natural numbers after 81858. Since

$$
52<\frac{81858}{1558}<53
$$

therefore $1558 \cdot 53=82574=m-4284.5$ is the first number, within the range of numbers, that is divisible by $1558$. The pair of $m-4284.5$ is $m+4284.5$, so the difference is $2 \cdot 4284.5=8569$. The difference of the next two suitable number pairs is clearly $8569-2 \cdot 1558=5453$, and finally the difference of the last number pair is $5453-3116=2337$.

2337 and 5453 have a single-digit divisor (3 and 7, respectively), but 8569 (=$11 \cdot 19 \cdot 41$) does not have a single-digit divisor, and thus the only solution is 82574 and $82574+8569=91143$.

II. solution: Every common divisor of the sum and one of the numbers also divides the other number, and thus is a divisor of the difference between the two numbers.

Therefore, $(173717,1558)=779=19 \cdot 41$ is also a divisor of the difference between the two numbers.

Thus, the difference is of the form $19 \cdot 41 \cdot k$, where $k$ cannot contain a single-digit prime factor.

Since according to the problem $1000 \leq 19 \cdot 41 \cdot k \leq 9999$, therefore

$$
\frac{1000}{779} \leq k \leq \frac{9999}{772}=12, \ldots
$$

or

$$
2 \leq k \leq 12
$$

Thus, only the value $k=11$ meets our conditions, and thus the difference is

$$
779 \cdot 11=8569
$$

If we denote the two sought numbers by $x$ and $y$, then

$$
\begin{aligned}
& x+y=173717 \\
& x-y=8569
\end{aligned}
$$

from which

$$
x=91143(=3 \cdot 13 \cdot 19 \cdot 41) \quad y=82574(=2 \cdot 19 \cdot 41 \cdot 53)
$$[^0]


[^0]:    ${ }^{1}$ In this case, the parentheses denote the greatest common divisor of the numbers separated by a comma within the parentheses.