Find out the greatest number of four digits which when divided by 5, 6, 8, 15 and 18 leaves remainder 2 in each case.

Options

A. 9362
B. 9722
C. 9822
D. 9972

Answer

To find the greatest four-digit number that leaves a remainder of 2 when divided by 5, 6, 8, 15, and 18, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( N \) such that:
- \( N \mod 5 = 2 \)
- \( N \mod 6 = 2 \)
- \( N \mod 8 = 2 \)
- \( N \mod 15 = 2 \)
- \( N \mod 18 = 2 \)

This means that \( N - 2 \) must be divisible by all these numbers.

### Step 2: Calculate the LCM
To find \( N - 2 \), we first need to calculate the Least Common Multiple (LCM) of the numbers 5, 6, 8, 15, and 18.

- The prime factorization of each number is:
- \( 5 = 5^1 \)
- \( 6 = 2^1 \times 3^1 \)
- \( 8 = 2^3 \)
- \( 15 = 3^1 \times 5^1 \)
- \( 18 = 2^1 \times 3^2 \)

- Now, take the highest power of each prime:
- For \( 2 \): \( 2^3 \) (from 8)
- For \( 3 \): \( 3^2 \) (from 18)
- For \( 5 \): \( 5^1 \) (from 5 and 15)

Thus, the LCM is:
\[ \text{LCM} = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360 \]

### Step 3: 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 divisible by 360.

### Step 4: Divide and Find the Largest Multiple
Now, we divide 9999 by 360:
\[ 9999 \div 360 \approx 27.775 \]
Taking the integer part, we have 27. Therefore, the largest multiple of 360 that is less than or equal to 9999 is:
\[ 27 \times 360 = 9720 \]

### Step 5: Add the Remainder
Since we need \( N \) such that \( N \mod 5 = 2 \), \( N \mod 6 = 2 \), etc., we add 2 to 9720:
\[ N = 9720 + 2 = 9722 \]

### Conclusion
The greatest four-digit number that leaves a remainder of 2 when divided by 5, 6, 8, 15, and 18 is **9722**.

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