How many numbers lie between 11 and 1111 which when divided by 9 leave a remainder of 6 and when divide by 21 leave a remainder of 12 ?

Answer

To solve the problem of finding how many numbers lie between 11 and 1111 that leave a remainder of 6 when divided by 9 and a remainder of 12 when divided by 21, we will follow these steps:

### Step 1: Set up the equations based on the conditions given.
Let the number be \( N \). According to the problem:
- When \( N \) is divided by 9, it leaves a remainder of 6. This can be expressed as:
\[ N = 9k + 6 \quad \text{(for some integer } k\text{)} \]
- When \( N \) is divided by 21, it leaves a remainder of 12. This can be expressed as:
\[ N = 21l + 12 \quad \text{(for some integer } l\text{)} \]

### Step 2: Set the two expressions for \( N \) equal to each other.
From the two equations, we have:
\[ 9k + 6 = 21l + 12 \]
Rearranging gives:
\[ 9k - 21l = 6 \]
Dividing the entire equation by 3 simplifies it to:
\[ 3k - 7l = 2 \]

### Step 3: Solve for \( k \) in terms of \( l \).
Rearranging the equation gives:
\[ 3k = 7l + 2 \implies k = \frac{7l + 2}{3} \]
For \( k \) to be an integer, \( 7l + 2 \) must be divisible by 3.

### Step 4: Find values of \( l \) that make \( k \) an integer.
To find valid values of \( l \), we can check the expression \( 7l + 2 \mod 3 \):
- \( 7 \mod 3 = 1 \), so:
\[ 7l + 2 \equiv l + 2 \mod 3 \]
We need \( l + 2 \equiv 0 \mod 3 \) which simplifies to:
\[ l \equiv 1 \mod 3 \]
Thus, \( l \) can take values of the form:
\[ l = 3m + 1 \quad \text{(for some integer } m\text{)} \]

### Step 5: Substitute \( l \) back to find \( k \).
Substituting \( l = 3m + 1 \) into the equation for \( k \):
\[ k = \frac{7(3m + 1) + 2}{3} = \frac{21m + 7 + 2}{3} = 7m + 3 \]

### Step 6: Substitute \( k \) back to find \( N \).
Now substitute \( k \) back into the equation for \( N \):
\[ N = 9k + 6 = 9(7m + 3) + 6 = 63m + 27 + 6 = 63m + 33 \]

### Step 7: Determine the range of \( N \).
We need to find values of \( N \) such that:
\[ 11 < N < 1111 \]
Substituting \( N = 63m + 33 \):
\[ 11 < 63m + 33 < 1111 \]
Subtracting 33 throughout:
\[ -22 < 63m < 1078 \]
Dividing by 63:
\[ -\frac{22}{63} < m < \frac{1078}{63} \]
Calculating the bounds:
\[ m > 0 \quad \text{(since } m \text{ must be a non-negative integer)} \]
Calculating the upper bound:
\[ \frac{1078}{63} \approx 17.11 \implies m \leq 17 \]

### Step 8: Count the integer values of \( m \).
The possible integer values for \( m \) are \( 0, 1, 2, \ldots, 17 \). This gives us:
\[ m = 0, 1, 2, \ldots, 17 \implies 18 \text{ values} \]

### Final Answer:
Thus, the total number of integers \( N \) that satisfy the conditions is **18**.

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