53(1139). I have thought of an integer not exceeding 1000. How can you find out what number I have thought of by asking no more than 10 questions, to which I will only answer “yes” or “no”?
Solution. When solving the problem, it is convenient to use a table of powers with base 2, as the number not exceeding 1000 is in the numerical interval $\left[2^{0}, 2^{10}\right)$. Questions should be asked in such a way that after each answer, the interval is reduced by half. (The found intervals will be marked on Figure 16.)

Let, for example, the number $300(x=300)$ be thought of.

Question 1. Is the number $x$ greater than 512? $\left(512=2^{9}\right)$.

Answer. No.

Question 2. Is the number $x$ greater than 256? $\left(256=2^{9}: 2=2^{8}\right)$. Answer. Yes.

Question 3. Is the number $x$ greater than 384? $\left(384=\frac{2^{8}+2^{9}}{2}\right)$. 

Answer. No.

Question 4. Is the number $x$ greater than 320? $\left(320=\frac{256+384}{2}\right)$. 

Answer. No.

Question 5. Is the number $x$ greater than 288? $\left(288=\frac{256+320}{2}\right)$. 

Answer. Yes.

Question 6. Is the number $x$ greater than 304? $\left(304=\frac{288+320}{2}\right)$. 

Answer. No.

Question 7. Is the number $x$ greater than 296? $\left(296=\frac{288+304}{2}\right)$. 

Answer. Yes.

Question 8. Is the number $x$ greater than 300? $\left(300=\frac{296+304}{2}\right)$. 

Answer. No.

Question 9. Is the number $x$ greater than 298? $\left(298=\frac{296+300}{2}\right)$. 

Answer. Yes.

Question 10. Is the number $x$ greater than 299? $\left(299=\frac{298+300}{2}\right)$. 

Answer. Yes.

Therefore, the number thought of is 300 (the only integer 300 is greater than 299 but not greater than 300).

$54(1140)$. In 5 years, the brother's age will be to the sister's age as 7:5. How old is each of them now if a year ago the brother was twice as old as the sister? Solution. I method. Let the sister be $x$ years old a year ago, and the brother be $2x$ years old. In 5 years, the sister will be $(x+6)$ years old, and the brother will be $(2x+6)$ years old. According to the problem, we have:

$$
\frac{2 x+6}{x+6}=\frac{7}{5}
$$

from which $x=4, 2x=8$, i.e., the sister is 5 years old now, and the brother is 9 years old.

II method. Let the brother be $7x$ years old in 5 years, and the sister be $5x$ years old. A year ago, they were $(7x-6)$ years and $(5x-6)$ years old, respectively. According to the problem, we have the equation:

$$
7 x-6=2(5 x-6)
$$

from which $x=2$.

The brother is 9 years old now $(7x-6+1)$, and the sister is 5 years old $(5x-6+1)$.

III method. Let the sister be $x$ years old now, and the brother be $y$ years old. According to the problem, we form the system of equations:

$$
\left\{\begin{array}{l}
\frac{y-1}{x-1}=2 \\
\frac{y+5}{x+5}=\frac{7}{5}
\end{array}\right.
$$

from which we find $x=5, y=9$.

Answer: the sister is 5 years old now, and the brother is 9 years old.

$55(1150)$. The first digit of a four-digit number is 7. If this digit is moved to the last place, the resulting number is 864 less than the original number. Find the original number.

Solution. I method. Let the required number be $\overline{7 a b c}$. According to the problem, we have:

$$
\overline{7 a b c}-\overline{a b c 7}=864
$$

If $\overline{a b c}=x$, then we have the equation:

$$
7000+x-(10 x+7)=864
$$

from which $x=681$, and the required number is 7681.

II method. The problem can be easily solved by selecting digits, rephrasing it as follows: find out what digits the letters represent, if each letter represents only a digit.

$$
\begin{array}{r}
7 a b c \\
-\frac{a b c 7}{864}
\end{array} \quad \text { or }+\frac{864}{7 a b c}
$$

from which $c=1, b=8, a=6$.

Answer: the required number is 7681.