Understanding the Problem: HCF and Algebraic Expressions

The question asks us to identify a number from the given options that is a guaranteed divisor of the highest common factor (HCF) of the two given algebraic expressions: \(36x^2 - 81y^2\) and \(81x^2 - 9y^2\). We are provided with a key piece of information: the HCF of \(x\) and \(y\) is 15.

Analyzing the Given Information

We are given that \(\text{HCF}(x, y) = 15\). This is a crucial starting point. When the HCF of two numbers is a specific value, say \(g\), it means the numbers can be expressed as multiples of \(g\). So, we can write:

• \(x = 15a\)
• \(y = 15b\)

where \(a\) and \(b\) are integers such that their HCF is 1 (\(\text{HCF}(a, b) = 1\)). This condition ensures that 15 is indeed the *highest* common factor of \(x\) and \(y\).

Factorizing the Algebraic Expressions

The given expressions are in the form of differences of squares. We can use the identity \(A^2 - B^2 = (A - B)(A + B)\) to factorize them.

First Expression: \(36x^2 - 81y^2\)

Recognize that \(36x^2 = (6x)^2\) and \(81y^2 = (9y)^2\).

\(36x^2 - 81y^2 = (6x)^2 - (9y)^2\)

Applying the difference of squares identity:

\(= (6x - 9y)(6x + 9y)\)

We can factor out a common factor of 3 from each bracket:

\(= 3(2x - 3y) \cdot 3(2x + 3y)\)

\(= 9(2x - 3y)(2x + 3y)\)

Second Expression: \(81x^2 - 9y^2\)

Recognize that \(81x^2 = (9x)^2\) and \(9y^2 = (3y)^2\).

\(81x^2 - 9y^2 = (9x)^2 - (3y)^2\)

Applying the difference of squares identity:

\(= (9x - 3y)(9x + 3y)\)

We can factor out a common factor of 3 from each bracket:

\(= 3(3x - y) \cdot 3(3x + y)\)

\(= 9(3x - y)(3x + y)\)

So, we are looking for the HCF of \(9(2x - 3y)(2x + 3y)\) and \(9(3x - y)(3x + y)\).

Substituting \(x = 15a\) and \(y = 15b\)

Now, let's substitute \(x = 15a\) and \(y = 15b\) into the factored expressions to see how they relate to \(a\) and \(b\).

First Expression: \(9(2x - 3y)(2x + 3y)\)

Substitute \(x=15a, y=15b\):

\(= 9(2(15a) - 3(15b))(2(15a) + 3(15b))\)

\(= 9(30a - 45b)(30a + 45b)\)

Factor out common factors from the brackets (15 from each):

\(= 9 \cdot 15(2a - 3b) \cdot 15(2a + 3b)\)

\(= 9 \cdot (15 \cdot 15) (2a - 3b)(2a + 3b)\)

\(= 9 \cdot 225 (4a^2 - 9b^2)\) (Using difference of squares again)

\(= 2025 (4a^2 - 9b^2)\)

Second Expression: \(9(3x - y)(3x + y)\)

Substitute \(x=15a, y=15b\):

\(= 9(3(15a) - (15b))(3(15a) + (15b))\)

\(= 9(45a - 15b)(45a + 15b)\)

Factor out common factors from the brackets (15 from each):

\(= 9 \cdot 15(3a - b) \cdot 15(3a + b)\)

\(= 9 \cdot (15 \cdot 15) (3a - b)(3a + b)\)

\(= 9 \cdot 225 (9a^2 - b^2)\) (Using difference of squares again)

\(= 2025 (9a^2 - b^2)\)

So the problem reduces to finding the HCF of \(2025(4a^2 - 9b^2)\) and \(2025(9a^2 - b^2)\), where \(\text{HCF}(a, b) = 1\).

Calculating the HCF of the Expressions

The HCF of the two expressions is:

\(\text{HCF}(2025(4a^2 - 9b^2), 2025(9a^2 - b^2))\)

Using the property \(\text{HCF}(kA, kB) = k \cdot \text{HCF}(A, B)\), we can factor out 2025:

\(= 2025 \cdot \text{HCF}(4a^2 - 9b^2, 9a^2 - b^2)\)

Let \(d = \text{HCF}(4a^2 - 9b^2, 9a^2 - b^2)\). Since \(d\) is the HCF, it must divide both \(4a^2 - 9b^2\) and \(9a^2 - b^2\).

If \(d\) divides two numbers, it must also divide any linear combination of those numbers (i.e., \(m \cdot \text{number}_1 + n \cdot \text{number}_2\) for integers \(m, n\)).

Consider the combination \(9 \cdot (9a^2 - b^2) - 1 \cdot (4a^2 - 9b^2)\):

\(9(9a^2 - b^2) - (4a^2 - 9b^2) = (81a^2 - 9b^2) - (4a^2 - 9b^2)\)

\(= 81a^2 - 9b^2 - 4a^2 + 9b^2 = 77a^2\)

So, \(d\) divides \(77a^2\).

Consider the combination \(9 \cdot (4a^2 - 9b^2) - 4 \cdot (9a^2 - b^2)\):

\(9(4a^2 - 9b^2) - 4(9a^2 - b^2) = (36a^2 - 81b^2) - (36a^2 - 4b^2)\)

\(= 36a^2 - 81b^2 - 36a^2 + 4b^2 = -77b^2\)

So, \(d\) divides \(-77b^2\), which means \(d\) divides \(77b^2\).

Since \(d\) divides both \(77a^2\) and \(77b^2\), it must divide their HCF: \(\text{HCF}(77a^2, 77b^2)\).

\(\text{HCF}(77a^2, 77b^2) = 77 \cdot \text{HCF}(a^2, b^2)\). Since \(\text{HCF}(a, b) = 1\), it follows that \(\text{HCF}(a^2, b^2) = 1\).

So, \(\text{HCF}(77a^2, 77b^2) = 77 \cdot 1 = 77\).

This means \(d\) is a divisor of 77. Possible values for \(d\) are the divisors of 77: 1, 7, 11, 77.

The HCF of the original expressions is \(2025 \cdot d\), where \(d\) is a divisor of 77.

Checking Divisibility by Options

We need to find which of the given options is guaranteed to divide \(2025 \cdot d\) for any \(d\) that is a divisor of 77. Let's find the prime factorization of 2025 and the options.

• \(2025 = 5 \times 405 = 5 \times 5 \times 81 = 5^2 \times 3^4 = 3^4 \times 5^2\)

Options:

• 135: \(135 = 3 \times 45 = 3 \times 9 \times 5 = 3 \times 3^2 \times 5 = 3^3 \times 5\)
• 120: \(120 = 12 \times 10 = (2^2 \times 3) \times (2 \times 5) = 2^3 \times 3 \times 5\)
• 180: \(180 = 18 \times 10 = (2 \times 3^2) \times (2 \times 5) = 2^2 \times 3^2 \times 5\)
• 90: \(90 = 9 \times 10 = 3^2 \times (2 \times 5) = 2 \times 3^2 \times 5\)

The HCF we found is \(2025 \cdot d = (3^4 \cdot 5^2) \cdot d\), where \(d \in \{1, 7, 11, 77\}\).

For the HCF to be divisible by an option, the prime factors of the option must be present in the HCF's factorization (\(3^4 \cdot 5^2 \cdot d\)) with at least the same powers.

• Option 1 (135 = \(3^3 \cdot 5\)): The HCF has \(3^4\) and \(5^2\). Since \(3^4\) contains \(3^3\) and \(5^2\) contains \(5^1\), \(2025 = 3^4 \cdot 5^2\) is divisible by \(135 = 3^3 \cdot 5\). Therefore, \(2025 \cdot d\) is always divisible by 135, regardless of the value of \(d\).
• Option 2 (120 = \(2^3 \cdot 3 \cdot 5\)): The number 2025 has no prime factor of 2. While \(d\) could be 7 or 11 or 77, none of these introduce a factor of 2. Thus, \(2025 \cdot d\) is never divisible by 120.
• Option 3 (180 = \(2^2 \cdot 3^2 \cdot 5\)): Similar to 120, 2025 has no factor of 2, and \(d\) does not introduce a factor of 2. Thus, \(2025 \cdot d\) is never divisible by 180.
• Option 4 (90 = \(2 \cdot 3^2 \cdot 5\)): Similar to 120 and 180, 2025 has no factor of 2, and \(d\) does not introduce a factor of 2. Thus, \(2025 \cdot d\) is never divisible by 90.

The only option that is guaranteed to divide the HCF \(2025 \cdot d\) is 135, because 135 is a factor of 2025 itself.

Conclusion

The HCF of \(36x^2 - 81y^2\) and \(81x^2 - 9y^2\) is found to be \(2025 \cdot d\), where \(d\) is a divisor of 77 (determined by the relationship between \(a\) and \(b\)). We determined that 2025 is divisible by 135. Therefore, the HCF \(2025 \cdot d\) will always be divisible by 135.

Revision Table: HCF and Algebraic Expressions

Additional Information: Working with HCF in Algebra

• When finding the HCF of algebraic expressions, always try to factorize them first. Look for common factors, differences of squares, perfect square trinomials, or grouping.
• If variables have a known HCF, like \(\text{HCF}(x, y) = g\), substituting \(x=ga\) and \(y=gb\) (with \(\text{HCF}(a,b)=1\)) is a powerful technique to simplify the problem in terms of coprime numbers.
• The HCF of two algebraic expressions is the product of all common factors raised to the lowest power they appear in either expression. For expressions involving variables, the HCF might also include variable terms.
• The concept that \(\text{HCF}(A, B)\) divides \(mA + nB\) is very useful for finding constraints on the HCF of terms that are difficult to factor directly, as demonstrated in finding the constraint on \(d\).