If N is natural number less than 100, then for how many values of N are the numbers `6N+1and15N+2` relatively prime?

Options

A. 16
B. 10
C. 33
D. All of these

Answer

To determine how many natural numbers \( N \) less than 100 make the numbers \( 6N + 1 \) and \( 15N + 2 \) relatively prime, we can follow these steps:

### Step 1: Understand the Problem
We need to find values of \( N \) such that the greatest common divisor (gcd) of \( 6N + 1 \) and \( 15N + 2 \) is 1, which means they are relatively prime.

**Hint:** Recall that two numbers are relatively prime if their gcd is 1.

### Step 2: Set Up the GCD Condition
We can use the property of gcd that states:
\[ \gcd(a, b) = \gcd(a, b - ka) \]
for any integer \( k \). We will apply this property to simplify our expressions.

**Hint:** Use the property of gcd to simplify the expressions.

### Step 3: Apply the GCD Property
Let \( a = 6N + 1 \) and \( b = 15N + 2 \). We can compute:
\[ \gcd(6N + 1, 15N + 2) = \gcd(6N + 1, (15N + 2) - 2(6N + 1)) \]
Calculating the second term:
\[ 15N + 2 - 2(6N + 1) = 15N + 2 - 12N - 2 = 3N \]
Thus, we have:
\[ \gcd(6N + 1, 15N + 2) = \gcd(6N + 1, 3N) \]

**Hint:** Simplifying the gcd can help us find the relationship between the two numbers.

### Step 4: Further Simplify the GCD
Now we can apply the gcd property again:
\[ \gcd(6N + 1, 3N) = \gcd(6N + 1 - 2(3N), 3N) = \gcd(6N + 1 - 6N, 3N) = \gcd(1, 3N) \]
Since \( \gcd(1, 3N) = 1 \), we find that \( 6N + 1 \) and \( 15N + 2 \) are relatively prime for all \( N \).

**Hint:** Remember that \( \gcd(1, x) = 1 \) for any integer \( x \).

### Step 5: Count the Valid Natural Numbers
Since \( N \) can be any natural number less than 100, we need to count all natural numbers from 1 to 99. There are 99 such numbers.

**Hint:** Count all natural numbers from 1 to 99 to find the total.

### Conclusion
Thus, the total number of values of \( N \) such that \( 6N + 1 \) and \( 15N + 2 \) are relatively prime is \( 99 \).

**Final Answer:** 99