78(1188). Check that the equality is correct:
a) $\sqrt{2 \frac{2}{3}}=2 \sqrt{\frac{2}{3}}$
b) $\sqrt{5 \frac{5}{24}}=5 \sqrt{\frac{5}{24}}$.

Specify the condition under which the observed pattern holds. Provide examples.
Solution. a) Method I. Transform the left side of the equation:

$$
\sqrt{2 \frac{2}{3}}=\sqrt{\frac{8}{3}}=\sqrt{\frac{4 \cdot 2}{3}}=2 \sqrt{\frac{2}{3}}
$$

Method II. Transform the right side of the equation:

$$
2 \sqrt{\frac{2}{3}}=\sqrt{\frac{8}{3}}=\sqrt{2 \frac{2}{3}}
$$

Method III. Square both sides of the equation:

$$
2 \frac{2}{3}=4 \cdot \frac{2}{3} ; 2 \frac{2}{3}=2 \frac{2}{3} .
$$

b) Method I. $\sqrt{5 \frac{5}{24}}=\sqrt{\frac{125}{24}}=\sqrt{\frac{25 \cdot 5}{24}}=5 \sqrt{\frac{5}{24}}$.

Method II. $5 \sqrt{\frac{5}{24}}=\sqrt{\frac{25 \cdot 5}{24}}=\sqrt{\frac{125}{24}}=\sqrt{5 \frac{5}{24}}$.

Method III. $5 \frac{5}{24}=25 \cdot \frac{5}{24} ; 5 \frac{5}{24}=5 \frac{5}{24}$.

Derive the condition under which

$$
\sqrt{k \frac{k}{n}}=k \sqrt{\frac{k}{n}}, \text { i.e., } \sqrt{\frac{k n+k}{n}}=k \sqrt{\frac{k}{n}}
$$

Method I. Square the obtained equation:

$$
\frac{k n+k}{n}=\frac{k^{3}}{n}
$$

from which $\frac{k(n+1)}{n}=\frac{k^{\prime}}{n}$, i.e., $n+1=k^{2}, k=\sqrt{n+1}$.

Thus, if $n+1=k^{2}$, then $\sqrt{k \frac{k}{n}}=k \sqrt{\frac{k}{n}}$. For example, if $n=15$, we get $k=4$ and $\sqrt{4 \frac{4}{15}}=4 \sqrt{\frac{4}{15}}$, if $n=80$, we get $k=9$ and $\sqrt{9 \frac{9}{80}}=9 \sqrt{\frac{9}{80}}$.

Method II. Let the integer part of the number $a$ be $x([a]=x)$, and the fractional part of the number $a$ be $y(\{a\}=y)$. Find the relationship between the integer and fractional parts of the number $a$, under which

$$
\sqrt{[a]+\{a\}}=[a] \cdot \sqrt{|a|} .
$$

## Square the equation (1):

$$
[a]+\{a\}=[a\}^{2} \cdot\{a\}
$$

from which $[a]=\{a\} \cdot\left([a]^{2}-1\right)$, i.e., $\{a\}=\frac{[a]}{[a]^{2}-1}$.

For example, if $[a]=6$, then $\{a\}=\frac{6}{36-1}=\frac{6}{35}$. We have $\sqrt{6 \frac{6}{35}}=$ $=6 \sqrt{\frac{6}{35}}$.

If $[a]=10$, then $\{a\}=\frac{10}{100-1}=\frac{10}{99}$. We have $\sqrt{10 \frac{10}{99}}=10 \sqrt{\frac{10}{99}}$.

$\triangle 79(1189)$. Find the value of the expression

$$
\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}
$$

Solution. Method I.

$$
\begin{aligned}
& \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{99+\sqrt{100}}}= \\
= & \frac{2-1}{\sqrt{2}+\sqrt{1}}+\frac{3-2}{\sqrt{3}+\sqrt{2}}+\frac{4-3}{\sqrt{4}+\sqrt{3}}+\ldots+\frac{100-99}{\sqrt{100}+\sqrt{99}}=
\end{aligned}
$$

$$
\begin{gathered}
=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\ldots+\sqrt{100}-\sqrt{99}= \\
=\sqrt{100}-\sqrt{1}=9
\end{gathered}
$$

Method II. By rationalizing the denominator of each fraction, we get that the denominator of each fraction will be equal to one, and the problem reduces to adding the numerators:

$$
\begin{gathered}
(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\ldots+\sqrt{100}-\sqrt{99}= \\
=\sqrt{100}-\sqrt{1}=9
\end{gathered}
$$

$80(1190)$. From the first hundred natural numbers, 51 numbers are taken arbitrarily. Prove that among the taken numbers, there are definitely two such numbers, one of which is a multiple of the other.

Solution. Any natural number can be represented in the form $2^{n} \cdot p$, where $n$ is a non-negative integer, and $p$ is an odd number. Represent each of the first hundred natural numbers in the form $2^{n} \cdot p$, where $n$ is a non-negative integer, and $p$ is an odd number (for odd numbers, $n=0$). For example: $15=$ $=2^{0} \cdot 15 ; 8=2^{3} \cdot 1 ; 24=2^{3} \cdot 3 ; 98=2 \cdot 49$.

Obviously, from two numbers $2^{n} \cdot p$ and $2^{n} \cdot p$, having the same factor $p$, one is a multiple of the other. Indeed, since the first hundred contains 50 odd numbers, the factor $p$ in the expression $2^{n} \cdot p$ for the numbers of the first hundred can take only 50 different values. Therefore, if we take 51 numbers from the first hundred, at least two of them will have the same factor $p$, and thus one of them is a multiple of the other.