61 (1189). A three-digit number $x$, divisible by 5, can be represented as the sum of the cube and the square of the same natural number. Find the number $x$.
Solution. Let $n$ be such a natural number that $x=n^{3}+n^{2}=n^{2}(n+1)$.

Since $x$ is divisible by 5 according to the condition, $x$ must end in 0 or 5. From the equality $x=n^{2}(n+1)$, we conclude that $x$ ends in 0. (The product $n^{2}(n+1)$ for $n=1,2,6,7$ ends in the digit 2, for $n=3,8$ - in the digit 6, and for $n=4,5,9,10$ in 0.) The condition of the problem is satisfied only by $n=5$ and $n=9$, since for $n=4$ the number $x$ is two-digit, and for $n \geqslant 10 x$ - a four-digit number.

Therefore, $x=5^{2}(5+1)=150$ or $x=9^{2}(9+1)=810$. Indeed, $150=5^{3}+5^{2} ; 810=9^{3}+9^{2}$.

Answer: $150 ; 810$.

$62(1190)$. Two different natural numbers were taken. These numbers were added, multiplied, the smaller was subtracted from the larger, and the larger was divided by the smaller. It turned out that the sum of all four results is 441. Find these numbers.

Solution. Let $a$ be the smaller number, and $n a$ be the larger. According to the condition, we have the equation

$$
(a+n a)+n a^{2}+(n a-a)+\frac{n a}{a}=441
$$

from which

$$
\begin{gathered}
2 n a+n a^{2}+n=441 \\
n\left(a^{2}+2 a+1\right)=441 \\
n(a+1)^{2}=441=3^{2} \cdot 7^{2}
\end{gathered}
$$

The problem reduces to solving the system of equations in natural numbers:

![](https://cdn.mathpix.com/cropped/2024_05_21_98932225530737b82071g-190.jpg?height=128&width=702&top_left_y=1344&top_left_x=671)

from which

$$
\left\{\begin{array} { l } 
{ n = 9 , } \\
{ a = 6 }
\end{array} \quad \text { or } \quad \left\{\begin{array}{l}
n=49 \\
a=2
\end{array}\right.\right.
$$

Answer: 6 and $54 ; 2$ and 98.

$\triangle$ 63(1191). Find two natural numbers, the difference of whose squares is 45.

Solution. According to the condition, $x^{2}-y^{2}=45$, where $x$ and $y$ are the required natural numbers. Since $x^{2}-y^{2}=45$, and $45=1 \cdot 45=3 \cdot 15=5 \cdot 9$, the problem reduces to solving the following systems of equations in natural numbers:

$$
\left\{\begin{array} { l } 
{ x - y = 1 ; } \\
{ x + y = 4 5 ; }
\end{array} \quad \left\{\begin{array} { l } 
{ x - y = 3 , } \\
{ x + y = 1 5 ; }
\end{array} \quad \left\{\begin{array}{l}
x-y=5 \\
x+y=9
\end{array}\right.\right.\right.
$$

Answer: $\{(23 ; 22),(9 ; 6),(7 ; 2)\}$.

$\triangle$ 64(1192). Prove that there does not exist a natural number which, upon the transposition of its first digit to the end of the number, increases by 5 times.

Proof. Since the number of digits of the natural number does not change, and it must increase by 5 times, the natural number in question must start with the digit 1. (In any other case, the product of the natural number by 5 contains one more digit than the natural number itself.) When this digit is moved to the end of the number, the resulting number ends in the digit 1. But such a number cannot be divisible by 5. The statement is proved.

$65(1175)^{\prime}$. Between the numbers $\frac{4}{7}$ and $\frac{5}{7}$, find any number that is the square of a rational number. How many solutions does this problem have?

Solution. Method 1. Let $\frac{a}{b}$ be a rational number satisfying the condition $\frac{4}{7}<\left(\frac{a}{b}\right)^{2}<\frac{5}{7}$. Since the function $x^2$ is monotonically increasing for $x>0$, we have $\sqrt{\frac{4}{7}}<\frac{a}{b}<\sqrt{\frac{5}{7}}$. Since $\sqrt{\frac{4}{7}} \approx 0.756$ and $\sqrt{\frac{5}{7}} \approx 0.840$, we need to find a rational number $\frac{a}{b}$ such that $0.756 < \frac{a}{b} < 0.840$. 

There are infinitely many such rational numbers, as the interval $(0.756, 0.840)$ contains infinitely many rational numbers. 

Answer: There are infinitely many solutions.

Method 2. Another way to solve this problem is discussed in $\$ 4$, p. 30.

Note. Another way to solve this problem is discussed in $\$ 4$, p. 30.

3 remark. Another way to solve this problem is discussed in $\$ 4$, p. 30.