\(\to\) The sum of the digits of 25a31 must be a multiple of 3. Therefore,

Since the given number is divisible by 3, the sum of its digits must be a multiple of 3.

For \(b=2\), the number becomes $$7a352$$. The sum of its digits simplifies to the condition that a+5 must be a multiple of 3. In this case, the possible values for the digit a are 1, 4, and 7.

For \(b=6\), the number becomes $$7a356$$. Evaluating the sum of its digits requires that \(a=3k \quad (k \in Z) \). In this case, the possible values for the digit a are 0, 3, 6, and 9. Thus, the sum of all distinct possible values for a is \( 1 + 4 +7+0+3+6+9=30\).

Divisibility by 5:

Since we must satisfy the condition $$5+7+4+a=3k \quad (k \in Z) $$ and the digit a must be even, the possible values for a are 2 and 8 (= 2 + 6). Therefore, the sum of these values is found to be 2 + 8 = 10.

Example:

Let \(r_0, r_1, r_2, \cdots, r_n\) represent the individual digits of an \((n+1)\)-digit number. For this integer to be divisible by 7, the following condition must hold for some integer \( k \in Z \):

$$(1\cdot r_0 + 3\cdot r_1+2\cdot r_2) – (1\cdot r_3 + 3\cdot r_4+2\cdot r_5)+ \cdots= 7k$$

Example:

Let’s verify that 17482591 is divisible by 7.

\[
\begin{array}{l l l l l l l l}
1 & 7 & 4 & 8 & 2 & 5 & 9 & 1\\
\downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \downarrow &\downarrow &\\
3 & 1 & 2 & 3 & 1 & 2 & 3 & 1\\
\end{array}
\]

\[
\Rightarrow (1 \cdot 1 + 3 \cdot 9 + 2 \cdot 5)- (1\cdot 2 + 3 \cdot 8 + 2 \cdot 4)+ (1\cdot 7 + 3 \cdot 1)= \\
38-34+10=14 \]

\[ 14 \text{ is a multiple of } 7, \text{ so } 17482591 \text{ is divisible by } 7. \]

To find the remainder of any integer when divided by 7, the same process is applied. The remainder obtained from the resulting value is equal to the remainder of the original integer.

Example:

Since subtracting 8 from the given number makes it divisible by 11, the number \(a3b73 – 8 = a3b65\) is exactly divisible by 11. Therefore,

\[
\begin{array}{l l l l l l l l}
a & 3 & b & 6 & 5 &\\
\downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \\
+ & – & + & – & +&\\
\end{array}
\]

\[
\begin{array}{l l}
\to (a+b+5)-(3+6) = 11\cdot k\\
\Rightarrow a+b-4=11\cdot k
\end{array}
\]

For \(k = -1\) \(\Rightarrow a+b= -7\) (The sum of two digits cannot be negative.)

For \(k = 0\) \(\Rightarrow a + b = 4\)

For \(k = 1\) \(\Rightarrow a + b = 15\)

For \(k = 2\) \(\Rightarrow a + b = 26\) (The sum of two single digits cannot be greater than 18.)

Accordingly, there are two distinct values that the sum of the digits a and b can take, which are 4 and 15.

Example:

For example, 300, 1425, 26750, and 1975 are divisible by 25.

Question 32:

\[
\begin{array}{c,@{\hspace{2cm}}c,c,c}
\begin{array}{c}
\quad &ab \;\;\\
\quad &\vdots\\
-\;\; \\
\hline
&\quad 2 \;\;
\end{array}
\begin{array}{|c}
\quad b \\
\hline
\quad 3
\\
\\
\end{array}
\end{array}
\]

Since \(60\cdot B\) is exactly divisible by 12, the remainder when the natural number A is divided by 12 is 10.

\({\textbf{Answer: D}}\)

Question 36:

\[
\begin{array}{c,@{\hspace{2cm}}c,c,c}
\begin{array}{c}
\quad &A \;\;\\
-\quad &\vdots \;\; \\
\hline
&\quad 2 \;\;
\end{array}
\begin{array}{|c}
\quad B+1 \\
\hline
\quad 3
\\
\\
\end{array}
\begin{array}{c}
\quad \text{and} \;\; \;\; \;\; \;\; \;\;
\end{array}
\begin{array}{C}
\quad &B \;\;\\
-\quad &\vdots \;\; \\
\hline
&\quad 1 \;\;
\end{array}
\begin{array}{|c}
\quad 5 \\
\hline
\quad C-1
\\
\\
\end{array}
\end{array} \]

For this number to be divisible by 3, the sum of its digits 3 + a + b must be a multiple of 3. Let’s find the largest and smallest values of a that satisfy these conditions while ensuring all digits are distinct:

Since the number 53ab leaves a remainder of 2 when divided by 4 and 5, subtracting 2 from this number makes it exactly divisible by both 4 and 5.

Given that the remainder when 53ab is divided by 5 is 2, the value of b can be either 2 or 7. Since the number minus 2 is divisible by both 4 and 5, this resulting number must be an even number. Therefore, the value of b can only be 2. In this case, the given number is 53a2 and the number minus 2 is 53a0.

Accordingly, for the number 53a0 to be divisible by 4, the values that can replace a are 0, 2, 4, 6, and 8, and the sum of these values is 20.

\({\textbf{Answer: E}}\)

The two numbers matching the required conditions are 984 and 104, so the difference between these two numbers is 880.

\({\textbf{Answer: B}}\)

Based on the information given in the question, subtracting 2 from the number means that 7a0b2 would be a number exactly divisible by 8. (Note: Alternatively, if 7a0b4 leaves a remainder of 2, then the number formed by its last three digits, 0b4, leaves a remainder of 2 when divided by 8, meaning \(0b4 – 2 = 0b2\) must be a multiple of 8).

For this number to be divisible by 8, the three-digit number 0b2 must be a multiple of 8. Therefore, b can be replaced by 3 or 7. Since the digits of the number must be distinct from each other, b cannot be 7 (as 7 is already used as the first digit). Thus, the number is of the form 7a034.

For this number to take its maximum value, we must choose a = 9. Accordingly, the sum of a + b matching the given conditions is 9 + 3 = 12.

\({\textbf{Answer: E}}\)

The five-digit number baa4b leaves a remainder of a when divided by 5. Given that this number is divisible by 4, what is the sum of the values that a can take?

\[ \text{A)} 12 \quad \text{B) } 10 \quad \text{C) } 7 \quad \text{D) } 6 \quad \text{E) } 4 \]

Solution:

Let the quotient of A divided by 111 be x.

\[\frac{A}{111} = x \in Z \Rightarrow A = 111 \cdot x\]

Since the number 111 is always divisible by 3, the number A = 111 $\cdot$ x is also always divisible by 3.

\({\textbf{Answer: B}}\)

Since the number xy is divisible by 9, the sum of its digits x + y is a multiple of 9. Accordingly, since the remainder when 7x6y is divided by 9 is equal to the remainder when its digit sum is divided by 9:

7 + 6 + x + y = 13 + (x + y) = 4 + 9 + (x + y). Therefore, the remainder when this number is divided by 9 is 4.

\({\textbf{Answer: A}}\)

Since the remainder when 2a55a is divided by 3 is 2:

\(2+a+5+5+a = 3\cdot k +2 \Rightarrow 12+2\cdot a =3\cdot k +2 \Rightarrow 10+2a= 3\cdot k \quad (k \in Z) \) holds.

Since the remainder when the given number is divided by 10 is 4, the units digit (c) of this number is 4. Since the number a25b384 is divisible by 11, applying the rule for divisibility by 11 gives:

\[
\begin{array}{l l l l l l l l}
a & 2 & 5 & b & 3 & 8 & 4 & \\
\downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \\
+ & – & + & – & + & – & + \\
\end{array}
\]

\[
\to (a+5+3+4)-(2+b+8)= 11\cdot k\Rightarrow a-b+2=11\cdot k \quad (k \in Z)
\] holds.

For \(k = -1 \quad \Rightarrow a-b= -13 \rightarrow \) impossible.

For \(k = 0 \quad \Rightarrow a-b= -2 \rightarrow \) possible.

For \(k = 1 \quad \Rightarrow a-b= 9 \rightarrow \) possible.

For \(k = 2 \quad \Rightarrow a-b= 20 \rightarrow \) impossible.

Since the number abcd8 leaves a remainder of 5 when divided by 11, subtracting 5 from this number (the number abcd3) makes it exactly divisible by 11.

The four-digit number \(aaa0\) is always divisible by which of the following?

\[ \text{A)} 8 \quad \text{B) } 12 \quad \text{C) } 15 \quad \text{D) } 18 \quad \text{E) } 20 \]

Solution: