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Question

If \(n\) is natural number less than 7, then what is the number of values of \(n\) for which \((12n+2)\) and \((8n+1)\) are relatively prime?

This question was previously asked in
CDS 2 2025 Maths Question Paper (14-Sep-2025)
The correct answer is
6

Determining Relative Primality for Given Expressions

The problem asks us to find the number of possible values for a natural number \(n\), where \(n\) is less than 7, such that the two expressions, \(12n+2\) and \(8n+1\), are relatively prime.

Understanding Relative Primality

Two integers are considered relatively prime (or coprime) if their greatest common divisor (GCD) is equal to 1. Mathematically, integers \(a\) and \(b\) are relatively prime if \(\text{gcd}(a, b) = 1\).

Identifying the Range of 'n'

The question specifies that \(n\) is a natural number and \(n < 7\). Natural numbers typically start from 1. Therefore, the possible values for \(n\) are:

  • \(n \in \{1, 2, 3, 4, 5, 6\}\)

We need to check for which of these values of \(n\) the expressions \(12n+2\) and \(8n+1\) are relatively prime.

Calculating the Greatest Common Divisor (GCD)

Let's find the GCD of the two expressions, \(12n+2\) and \(8n+1\). We can use the properties of GCD. Let \(d = \text{gcd}(12n+2, 8n+1)\).

According to the properties of GCD, \(d\) must divide any integer linear combination of \(12n+2\) and \(8n+1\). We can manipulate these expressions to eliminate \(n\).

Consider the multiples of the expressions:

  • Multiply \((8n+1)\) by 2: \(2 \times (8n+1) = 16n+2\).
  • Multiply \((12n+2)\) by 1: \(1 \times (12n+2) = 12n+2\).

Alternatively, we can use the property \(\text{gcd}(a, b) = \text{gcd}(a, b-ka)\). Let \(a = 12n+2\) and \(b = 8n+1\). We aim to simplify the expression.

Using the Euclidean algorithm property, \(\text{gcd}(a, b) = \text{gcd}(a - k \cdot b, b)\). Let's try to eliminate the highest power term (\(n\)).

We can write:

\(\text{gcd}(12n+2, 8n+1)\)

\(= \text{gcd}(12n+2 - 1 \cdot (8n+1), 8n+1)\)

\(= \text{gcd}(12n+2 - 8n - 1, 8n+1)\)

\(= \text{gcd}(4n+1, 8n+1)\)

Now, apply the property again:

\(\text{gcd}(4n+1, 8n+1) = \text{gcd}(4n+1, 8n+1 - 2 \cdot (4n+1))\)

\(= \text{gcd}(4n+1, 8n+1 - 8n - 2)\)

\(= \text{gcd}(4n+1, -1)\)

The greatest common divisor of any integer \(x\) and \(-1\) is always 1 (since GCD is defined as a positive integer). Therefore:

\(\text{gcd}(4n+1, -1) = 1\).

This calculation shows that \(\text{gcd}(12n+2, 8n+1) = 1\) for all integer values of \(n\). This means the expressions \(12n+2\) and \(8n+1\) are always relatively prime, regardless of the value of \(n\).

Counting the Values of 'n'

Since the expressions are relatively prime for all integer values of \(n\), they will certainly be relatively prime for all the allowed values of \(n\) in the set \(\{1, 2, 3, 4, 5, 6\}\).

The number of possible values for \(n\) is the count of elements in this set, which is 6.

Conclusion

There are 6 values of \(n\) (namely 1, 2, 3, 4, 5, and 6) for which \(n\) is a natural number less than 7 and \((12n+2)\) and \((8n+1)\) are relatively prime.

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