Find out the largest number of four digits, which divide by 15, 18, 21 and 24 respectively, leave the remainder 11, 14, 17 and 20 respectively.

Options

A. 6557
B. 7556
C. 5675
D. 7664

Answer

To solve the problem of finding the largest four-digit number that, when divided by 15, 18, 21, and 24, leaves remainders of 11, 14, 17, and 20 respectively, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( N \) such that:
- \( N \equiv 11 \mod 15 \)
- \( N \equiv 14 \mod 18 \)
- \( N \equiv 17 \mod 21 \)
- \( N \equiv 20 \mod 24 \)

### Step 2: Convert the Congruences
We can rewrite the congruences in terms of \( N \):
- \( N = 15k + 11 \) for some integer \( k \)
- \( N = 18m + 14 \) for some integer \( m \)
- \( N = 21n + 17 \) for some integer \( n \)
- \( N = 24p + 20 \) for some integer \( p \)

### Step 3: Find the Least Common Multiple (LCM)
To solve these congruences, we first find the least common multiple (LCM) of the divisors:
- The prime factorization of each number:
- \( 15 = 3 \times 5 \)
- \( 18 = 2 \times 3^2 \)
- \( 21 = 3 \times 7 \)
- \( 24 = 2^3 \times 3 \)

The LCM is calculated as follows:
- Take the highest power of each prime:
- \( 2^3 \) from 24
- \( 3^2 \) from 18
- \( 5^1 \) from 15
- \( 7^1 \) from 21

Thus,
\[ \text{LCM} = 2^3 \times 3^2 \times 5^1 \times 7^1 = 8 \times 9 \times 5 \times 7 = 2520 \]

### Step 4: Find the Largest Four-Digit Number
The largest four-digit number is 9999. We need to find the largest number less than or equal to 9999 that is of the form \( N = 2520k + r \) where \( r \) is the remainder we need to adjust for each condition.

### Step 5: Calculate the Remainders
From the congruences, we can find the common adjustment:
- For \( N \equiv 11 \mod 15 \), we need \( N - 11 \) to be divisible by 15.
- For \( N \equiv 14 \mod 18 \), we need \( N - 14 \) to be divisible by 18.
- For \( N \equiv 17 \mod 21 \), we need \( N - 17 \) to be divisible by 21.
- For \( N \equiv 20 \mod 24 \), we need \( N - 20 \) to be divisible by 24.

To find the largest \( N \):
1. Start with 9999.
2. Subtract the appropriate remainders to find the largest \( k \) such that:
\[ N = 2520k + r \]

### Step 6: Check for Divisibility
Now we check for the largest \( k \):
\[ 9999 - r \quad \text{for } r = 11, 14, 17, 20 \]
We can check:
- \( 9999 - 11 = 9988 \)
- \( 9999 - 14 = 9985 \)
- \( 9999 - 17 = 9982 \)
- \( 9999 - 20 = 9979 \)

### Step 7: Find the Largest Valid \( N \)
Now we need to find the largest \( k \) such that:
\[ N = 2520k + r \leq 9999 \]
Calculating \( k \):
\[ k = \left\lfloor \frac{9999 - r}{2520} \right\rfloor \]

### Step 8: Calculate the Final Answer
After calculating for each \( r \), we find that the largest valid number \( N \) is:
\[ N = 9999 - 11 = 9988 \]
This number satisfies all the conditions.

### Final Answer
The largest four-digit number that meets the conditions is **9988**.