Institution: Bangladesh Development Bank Exam: Bangladesh Development Bank Ltd - Senior Officer - 24.11.2017
Topic: Arithmetic Mathematics (পাটিগণিত), Subject: Mathematics (সাধারণ গণিত)

Best Answer Explained in Below-:

To solve the problem of finding integers among the first thousand natural numbers that leave a remainder of 4 when divided by 7 and a remainder of 9 when divided by 11, we can express this using congruences:

1. \( x \equiv 4 \mod 7 \)
2. \( x \equiv 9 \mod 11 \)

We will use the method of substitution to find a common solution. From the first congruence, we can express \( x \) as:

\[ x = 7k + 4 \]

for some integer \( k \). Next, we substitute this expression into the second congruence:

\[ 7k + 4 \equiv 9 \mod 11 \]

Subtracting 4 from both sides gives:

\[ 7k \equiv 5 \mod 11 \]

To solve for \( k \), we need the multiplicative inverse of 7 modulo 11. The inverse can be found by trying values and checking:

\[ 7 \times 8 = 56 \equiv 1 \mod 11 \]

Thus, multiplying both sides of \( 7k \equiv 5 \) by 8 gives:

\[ k \equiv 8 \times 5 = 40 \equiv 7 \mod 11 \]

This indicates that:

\[ k = 11m + 7 \]

for some integer \( m \). Substituting back into \( x \):

\[ x = 7(11m + 7) + 4 = 77m + 49 + 4 = 77m + 53 \]

Now, we want \( x \) values under 1000:

\[ 77m + 53 \leq 1000 \]

This simplifies to:

\[ 77m \leq 947 \]
\[ m \leq 12.29 \]

Thus, \( m \) can take integer values from 0 to 12, giving a total of 13 possibilities (0 through 12).

Therefore, the answer is B) 13.