Match the column
Consider the number = N 123X43Y where x & y are digits `0leXle9and0leyle9.` Now answer the following

Options

A. `{:("If N is divisible by 2, then the sum of all possible",(p),57):}`
B. `{:("If N is divisible by 3, then the sum of all possiblevale of x+y will be",(q),17):}`
C. `{:("If N is divisible by 6, then number of orded pairs (x,y) can be",(r),20):}`
D. `{:("If N is divisiable by 9, then the sum of possible value of ",(s),19):}`

Answer

To solve the problem regarding the number \( N = 123X43Y \), where \( X \) and \( Y \) are digits ranging from 0 to 9, we will analyze the conditions for divisibility by 2, 3, 6, and 9, and find the corresponding sums or counts as required.

### Step-by-Step Solution:

1. **Divisibility by 2**:
- A number is divisible by 2 if its last digit is even.
- Here, the last digit is \( Y \). Therefore, \( Y \) can be \( 0, 2, 4, 6, \) or \( 8 \).
- The sum of these possible values of \( Y \) is:
\[ 0 + 2 + 4 + 6 + 8 = 20 \]
- **Match**: If \( N \) is divisible by 2, then the sum of all possible values of \( Y \) is \( 20 \) (matches with option R).

2. **Divisibility by 3**:
- A number is divisible by 3 if the sum of its digits is divisible by 3.
- The sum of the digits in \( N \) is:
\[ 1 + 2 + 3 + X + 4 + 3 + Y = 13 + X + Y \]
- For \( N \) to be divisible by 3, \( 13 + X + Y \) must be divisible by 3.
- The possible values for \( X + Y \) can be calculated as follows:
- The minimum value of \( X + Y \) is \( 0 \) (when \( X = 0, Y = 0 \)).
- The maximum value of \( X + Y \) is \( 18 \) (when \( X = 9, Y = 9 \)).
- The possible sums \( 13 + X + Y \) that are divisible by 3 can be \( 15, 18, 21, 24, 27 \) (i.e., \( X + Y = 2, 5, 8, 11, 14, 17 \)).
- The sum of these values is:
\[ 2 + 5 + 8 + 11 + 14 + 17 = 57 \]
- **Match**: If \( N \) is divisible by 3, then the sum of all possible values of \( X + Y \) is \( 57 \) (matches with option P).

3. **Divisibility by 6**:
- A number is divisible by 6 if it is divisible by both 2 and 3.
- We already established that \( Y \) must be even for divisibility by 2.
- We will now count the ordered pairs \( (X, Y) \) that satisfy both conditions.
- We will check each even \( Y \) value:
- For \( Y = 0 \): \( 13 + X \) must be divisible by 3. Possible \( X \) values: \( 0, 3, 6, 9 \) (4 pairs).
- For \( Y = 2 \): \( 15 + X \) must be divisible by 3. Possible \( X \) values: \( 0, 3, 6, 9 \) (4 pairs).
- For \( Y = 4 \): \( 17 + X \) must be divisible by 3. Possible \( X \) values: \( 1, 4, 7 \) (3 pairs).
- For \( Y = 6 \): \( 19 + X \) must be divisible by 3. Possible \( X \) values: \( 0, 3, 6, 9 \) (4 pairs).
- For \( Y = 8 \): \( 21 + X \) must be divisible by 3. Possible \( X \) values: \( 0, 3, 6, 9 \) (4 pairs).
- Total ordered pairs: \( 4 + 4 + 3 + 4 + 4 = 19 \).
- **Match**: If \( N \) is divisible by 6, then the number of ordered pairs \( (X, Y) \) is \( 19 \) (matches with option Q).

4. **Divisibility by 9**:
- A number is divisible by 9 if the sum of its digits is divisible by 9.
- The sum of the digits is \( 13 + X + Y \).
- We need \( 13 + X + Y \) to be divisible by 9.
- Possible values of \( X + Y \) that satisfy \( 13 + X + Y \equiv 0 \mod 9 \):
- \( X + Y = 5 \) (i.e., \( 13 + 5 = 18 \)).
- \( X + Y = 14 \) (i.e., \( 13 + 14 = 27 \)).
- The sum of these possible values is:
\[ 5 + 14 = 19 \]
- **Match**: If \( N \) is divisible by 9, then the sum of possible values of \( X + Y \) is \( 19 \) (matches with option S).

### Summary of Matches:
- If \( N \) is divisible by 2, then the sum is \( 20 \) (R).
- If \( N \) is divisible by 3, then the sum is \( 57 \) (P).
- If \( N \) is divisible by 6, then the number of ordered pairs is \( 19 \) (Q).
- If \( N \) is divisible by 9, then the sum is \( 19 \) (S).