Is there a base of a number system for which the number represented as "72" is divisible by the number represented as "275"?
I. solution. Suppose that the system in question exists and denote its base by $b$. Then

$$
\frac{5 b^{2}+7 b+2}{2 b^{2}+7 b+5}=q
$$

where $q$ is a natural number. Due to the use of the digit 7, it is clear that $b \geq 8$. The two expressions can be factored similarly, leading to

$$
q=\frac{(5 b+2)(b+1)}{(2 b+5)(b+1)}=\frac{5 b+2}{2 b+5}=2+\frac{b-8}{2 b+5}
$$

Here, the value of the last fractional expression is an integer and non-negative, since the numerator is non-negative and the denominator is positive. However, it cannot be a positive integer, as the denominator is greater than the numerator for the allowed values of $b$:

$$
(2 b+5)-(b-8)=b+13 \geqq 21>0
$$

Therefore, the value of the fraction can only be 0, and from this $b=8, q=2$. (Indeed, ${ }^{8)} 275={ }^{10)} 189,{ }^{8} 572={ }^{10)} 378=2 .{ }^{10)} 189$. )

II. solution. From equation (1), the following quadratic equation in $b$ is obtained

$$
(5-2 q) b^{2}+7(1-q) b+(2-5 q)=0
$$

where $q$ must be chosen as a natural number such that the equation has an integer root greater than 7. From (2),

$$
b=\frac{-7+7 q \pm \sqrt{49(1-q)^{2}-4(5-2 q)(2-5 q)}}{2(5-2 q)}=\frac{-7+7 q \pm \sqrt{9 q^{2}+18 q+9}}{2(5-2 q)}=\left\{\begin{array}{c}
\frac{5 q-2}{5-2 q} \\
-1
\end{array}\right.
$$

Thus, we need to choose such a $q$ that

$$
\frac{5 q-2}{5-2 q} \geq 8, \quad \frac{5 q-2}{5-2 q}-8=\frac{21(q-2)}{5-2 q} \geq 0
$$

Here, the numerator is negative for $q>5 / 2$. Thus, the fraction is defined and non-negative for $2 \leq q<5 / 2$, so the condition is only satisfied by the integer $q=2$. In this case, the base value $b=8$ is obtained.