Find the smallest number which when increased by 17 is exactly divisible by both 468 and 520.

Options

A. `5663`
B. `6663`
C. `4663`
D. `2663`

Explanation

To solve the problem of finding the smallest number which, when increased by 17, is exactly divisible by both 468 and 520, we can follow these steps:

### Step 1: Find the LCM of 468 and 520

To find the least common multiple (LCM), we can use the prime factorization method.

1. **Prime Factorization of 468:**
- Divide by 2: \( 468 \div 2 = 234 \)
- Divide by 2: \( 234 \div 2 = 117 \)
- Divide by 3: \( 117 \div 3 = 39 \)
- Divide by 3: \( 39 \div 3 = 13 \)
- Divide by 13: \( 13 \div 13 = 1 \)

So, the prime factorization of 468 is:
\[ 468 = 2^2 \times 3^2 \times 13^1 \]

2. **Prime Factorization of 520:**
- Divide by 2: \( 520 \div 2 = 260 \)
- Divide by 2: \( 260 \div 2 = 130 \)
- Divide by 2: \( 130 \div 2 = 65 \)
- Divide by 5: \( 65 \div 5 = 13 \)
- Divide by 13: \( 13 \div 13 = 1 \)

So, the prime factorization of 520 is:
\[ 520 = 2^3 \times 5^1 \times 13^1 \]

3. **Finding the LCM:**
- Take the highest power of each prime factor:
- For 2: \( 2^3 \)
- For 3: \( 3^2 \)
- For 5: \( 5^1 \)
- For 13: \( 13^1 \)

Therefore, the LCM is:
\[ LCM = 2^3 \times 3^2 \times 5^1 \times 13^1 = 8 \times 9 \times 5 \times 13 \]

Calculating this step by step:
- \( 8 \times 9 = 72 \)
- \( 72 \times 5 = 360 \)
- \( 360 \times 13 = 4680 \)

So, \( LCM(468, 520) = 4680 \).

### Step 2: Find the smallest number

Now, we need to find the smallest number \( x \) such that \( x + 17 \) is divisible by both 468 and 520. This means:
\[ x + 17 = 4680k \quad \text{for some integer } k \]
Thus,
\[ x = 4680k - 17 \]

To find the smallest positive \( x \), we set \( k = 1 \):
\[ x = 4680 \cdot 1 - 17 = 4680 - 17 = 4663 \]

### Final Answer:

The smallest number which when increased by 17 is exactly divisible by both 468 and 520 is **4663**.

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