64(1168). Can: a) the sum of five consecutive natural numbers be a prime number; b) the sum of the squares of five consecutive natural numbers be a prime number?
Solution. a) Let's check on specific examples whether the sum of five consecutive natural numbers is a prime or composite number: $1+2+3+4+5=15 ; 2+3+4+5+6=$ $=20 ; 3+4+5+6+7=25 ; \ldots ; 10+11+12+13+14=60$.

The numbers $15,20,25,60$ are composite numbers. We get the hypothesis: the sum of five consecutive natural numbers is a composite number. Let's prove it. We have:

$$
A=n+(n+1)+(n+2)+(n+3)+(n+4)=5 n+10=5(n+2) .
$$

Since by condition $n \in \boldsymbol{N}$, then $n+2 \in \boldsymbol{N}$ and $5(n+2) \in \boldsymbol{N}$, and $5(n+2)>5$. Therefore, the specified sum is divisible by 5, but not equal to 5, i.e., $A$ is a composite number.

b) Let's check on specific examples that the sum of the squares of five consecutive natural numbers is a composite number:

$$
\begin{gathered}
1^{2}+2^{2}+3^{2}+4^{2}+5^{2}=55,2^{2}+3^{2}+4^{2}+5^{2}+6^{2}=90 \\
3^{2}+4^{2}+5^{2}+6^{2}+7^{2}=135,10^{2}+11^{2}+12^{2}+13^{2}+14^{2}=730
\end{gathered}
$$

The numbers $55,90,135,730$ are composite.

To prove this fact in general, let's write five consecutive natural numbers as: $n-2, n-1, n, n+1$, $n+2$, where $n$ is any natural number greater than 2. Then we have:

$$
\begin{gathered}
B=(n-2)^{2}+(n-1)^{2}+n^{2}+(n+1)^{2}+(n+2)^{2}= \\
=5 n^{2}+10=5\left(n^{2}+2\right) .
\end{gathered}
$$

Since $n \in \boldsymbol{N}$, then $5\left(n^{2}+2\right) \in \boldsymbol{N}$ and $5\left(n^{2}+2\right)>5$. Therefore, the number $B$ is divisible by 5, but not equal to 5, i.e., $B$ is a composite number.

$\square$ 65(1181). If the thought number is divided by the sum of its digits, the quotient is 4 and the remainder is 3. If the thought number is reduced by twice the sum of its digits, the result is 25. What is the thought number?

$\mathrm{Pe}$ solution. Let the desired number be $\overline{a b}=10 a+b$. According to the problem, we form a system of equations:

$$
\left\{\begin{array}{c}
10 a+b=4(a+b)+3 \\
10 a+b-2(a+b)=25
\end{array}\right.
$$

which, after simplification, has the form:

$$
\left\{\begin{array}{l}
2 a-b=1 \\
8 a-b=25
\end{array}\right.
$$

from which $a=4, b=7$.

Answer: the desired number is 47.