How many natural numbers less than 500, when divided by 35 and 49, will leave a remainder of 30 in each case

Options

A. 4
B. 3
C. 1
D. 2

Answer

To solve the problem of finding how many natural numbers less than 500 leave a remainder of 30 when divided by both 35 and 49, we can follow these steps:

### Step 1: Understand the Problem
We need to find natural numbers \( n \) such that:
- \( n \equiv 30 \mod 35 \)
- \( n \equiv 30 \mod 49 \)

### Step 2: Rewrite the Congruences
From the congruences, we can express \( n \) in the following form:
- \( n = 35k + 30 \) for some integer \( k \)
- \( n = 49m + 30 \) for some integer \( m \)

### Step 3: Find the LCM of 35 and 49
To solve this problem, we need to find the least common multiple (LCM) of 35 and 49.

- The prime factorization of 35 is \( 5 \times 7 \).
- The prime factorization of 49 is \( 7^2 \).

The LCM is calculated by taking the highest power of each prime:
- LCM = \( 5^1 \times 7^2 = 5 \times 49 = 245 \).

### Step 4: General Form of the Solution
The numbers \( n \) that satisfy both conditions can be expressed as:
\[ n = 245k + 30 \]
where \( k \) is a non-negative integer.

### Step 5: Find Values of \( n \) Less Than 500
Now we need to find the values of \( k \) such that \( n < 500 \):
\[ 245k + 30 < 500 \]
Subtracting 30 from both sides gives:
\[ 245k < 470 \]
Dividing both sides by 245 gives:
\[ k < \frac{470}{245} \approx 1.918 \]

Since \( k \) must be a non-negative integer, the possible values for \( k \) are 0 and 1.

### Step 6: Calculate Corresponding Values of \( n \)
Now we calculate \( n \) for these values of \( k \):
- For \( k = 0 \):
\[ n = 245 \times 0 + 30 = 30 \]
- For \( k = 1 \):
\[ n = 245 \times 1 + 30 = 275 \]

### Step 7: Check for Validity
Both values \( 30 \) and \( 275 \) are less than 500.

### Conclusion
The natural numbers less than 500 that leave a remainder of 30 when divided by both 35 and 49 are \( 30 \) and \( 275 \). Thus, there are **2 natural numbers** that satisfy the condition.

### Final Answer
The answer is **2**.

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