Consider the number ` N= 123X43Y` where x & y are digits `0leXle9and0leYle9.` Now answer the following
`{:("If N is divisible by 6, then number of orded pairs (x,y) can be":}`

Options

A. 57
B. 17
C. 20
D. 19

Answer

To determine the number of ordered pairs \((x, y)\) such that the number \(N = 123X43Y\) is divisible by 6, we need to check two conditions:

1. **Divisibility by 2**: The last digit \(Y\) must be even.
2. **Divisibility by 3**: The sum of the digits of \(N\) must be divisible by 3.

### Step 1: Check for divisibility by 2
For \(N\) to be divisible by 2, \(Y\) must be one of the even digits: \(0, 2, 4, 6, 8\).

This gives us 5 possible values for \(Y\).

### Step 2: Check for divisibility by 3
Next, we need to find the sum of the digits of \(N\):
\[ 1 + 2 + 3 + X + 4 + 3 + Y = 13 + X + Y \]
This sum \(13 + X + Y\) must be divisible by 3.

### Step 3: Determine the values of \(X\) and \(Y\)
Now, we will analyze the values of \(Y\) (which can be \(0, 2, 4, 6, 8\)) and find corresponding values of \(X\) that make \(13 + X + Y\) divisible by 3.

1. **If \(Y = 0\)**:
\[ 13 + X + 0 = 13 + X \equiv 0 \mod 3 \]
\(13 \equiv 1 \mod 3\), so \(X \equiv 2 \mod 3\). Possible values for \(X\) are \(2, 5, 8\) (3 values).

2. **If \(Y = 2\)**:
\[ 13 + X + 2 = 15 + X \equiv 0 \mod 3 \]
\(15 \equiv 0 \mod 3\), so \(X \equiv 0 \mod 3\). Possible values for \(X\) are \(0, 3, 6, 9\) (4 values).

3. **If \(Y = 4\)**:
\[ 13 + X + 4 = 17 + X \equiv 0 \mod 3 \]
\(17 \equiv 2 \mod 3\), so \(X \equiv 1 \mod 3\). Possible values for \(X\) are \(1, 4, 7\) (3 values).

4. **If \(Y = 6\)**:
\[ 13 + X + 6 = 19 + X \equiv 0 \mod 3 \]
\(19 \equiv 1 \mod 3\), so \(X \equiv 2 \mod 3\). Possible values for \(X\) are \(2, 5, 8\) (3 values).

5. **If \(Y = 8\)**:
\[ 13 + X + 8 = 21 + X \equiv 0 \mod 3 \]
\(21 \equiv 0 \mod 3\), so \(X \equiv 0 \mod 3\). Possible values for \(X\) are \(0, 3, 6, 9\) (4 values).

### Step 4: Count the total ordered pairs \((x, y)\)
Now, we can sum the number of valid pairs for each case of \(Y\):

- For \(Y = 0\): 3 values of \(X\)
- For \(Y = 2\): 4 values of \(X\)
- For \(Y = 4\): 3 values of \(X\)
- For \(Y = 6\): 3 values of \(X\)
- For \(Y = 8\): 4 values of \(X\)

Total pairs:
\[ 3 + 4 + 3 + 3 + 4 = 17 \]

### Final Answer
The number of ordered pairs \((x, y)\) such that \(N\) is divisible by 6 is **17**.

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