Find the greatest four - digits number which when divided by 4,7 and 13 leaves a remainder 3 in each case.

Options

A. `9831`
B. `7831`
C. `8831`
D. `8821`

Explanation

To solve the problem of finding the greatest four-digit number that leaves a remainder of 3 when divided by 4, 7, and 13, we can follow these steps:

### Step 1: Identify the greatest four-digit number
The greatest four-digit number is 9999.

### Step 2: Determine the LCM of the divisors
To find a number that leaves a remainder of 3 when divided by 4, 7, and 13, we first need to find the least common multiple (LCM) of these numbers.

- The prime factorization of the numbers is:
- 4 = \(2^2\)
- 7 = \(7^1\)
- 13 = \(13^1\)

The LCM is calculated by taking the highest power of each prime factor:
- LCM(4, 7, 13) = \(2^2 \times 7^1 \times 13^1 = 4 \times 7 \times 13\)

Calculating this:
- \(4 \times 7 = 28\)
- \(28 \times 13 = 364\)

So, LCM(4, 7, 13) = 364.

### Step 3: Adjust the greatest four-digit number
Since we want a number that leaves a remainder of 3, we can express our desired number \(N\) in the form:
\[ N = k \times 364 + 3 \]
where \(k\) is an integer.

To find the largest \(N\) that is a four-digit number, we set:
\[ N \leq 9999 \]
Thus,
\[ k \times 364 + 3 \leq 9999 \]
Subtracting 3 from both sides gives:
\[ k \times 364 \leq 9996 \]
Dividing both sides by 364 gives:
\[ k \leq \frac{9996}{364} \approx 27.5 \]

Since \(k\) must be an integer, we take \(k = 27\).

### Step 4: Calculate the greatest number
Now we can calculate \(N\):
\[ N = 27 \times 364 + 3 \]

Calculating \(27 \times 364\):
\[ 27 \times 364 = 9828 \]

Adding 3:
\[ N = 9828 + 3 = 9831 \]

### Final Result
The greatest four-digit number that leaves a remainder of 3 when divided by 4, 7, and 13 is **9831**.

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