Question

If \(n\) is a natural number then \(9^{2n} - 4^{2n}\) is always divisible by :

• A. 5
• B. 13
• C. both 5 and 13
• D. none of these

Hint:

1. Rewrite the expression: \(9^{2n} - 4^{2n} = (9^2)^n - (4^2)^n = 81^n - 16^n\). 2. Use the property: \(a^n - b^n\) is always divisible by \(a-b\). Here, \(a=81, b=16\). So, \(a-b = 81-16 = 65\). 3. Check divisors of 65: \(65 = 5 \times 13\). Since the expression is divisible by 65, it must be divisible by 5 and by 13. Alternatively, test for \(n=1\): \(9^2 - 4^2 = 81 - 16 = 65\). \(65\) is divisible by 5 and 13.

Answer 1

The Correct Option is C

Concept: This problem involves testing divisibility of an algebraic expression. We can use algebraic identities or test small values of \(n\). The relevant algebraic identity is the difference of squares: \(a^2 - b^2 = (a-b)(a+b)\). More generally, \(a^k - b^k\) is divisible by \((a-b)\) for any positive integer \(k\), and divisible by \((a+b)\) if \(k\) is an even positive integer. Step 1: Rewrite the expression The given expression is \(9^{2n} - 4^{2n}\). We can rewrite \(9^{2n}\) as \((9^2)^n = 81^n\). We can rewrite \(4^{2n}\) as \((4^2)^n = 16^n\). So the expression becomes \(81^n - 16^n\). Using the property that \(a^n - b^n\) is always divisible by \((a-b)\): Here, \(a=81\) and \(b=16\). So, \(81^n - 16^n\) is always divisible by \((81-16)\). \(81 - 16 = 65\). The divisors of 65 are 1, 5, 13, 65. Therefore, \(81^n - 16^n\) is always divisible by 5 and by 13. Alternative Method using Difference of Squares repeatedly (if n allows) Let \(x = 9^n\) and \(y = 4^n\). Then the expression is \( (9^n)^2 - (4^n)^2 = x^2 - y^2 = (x-y)(x+y) \). So, \(9^{2n} - 4^{2n} = (9^n - 4^n)(9^n + 4^n)\). Step 2: Test with small values of \(n\) (since \(n\) is a natural number, \(n=1, 2, 3, \ldots\)) Case 1: Let \(n=1\). \(9^{2(1)} - 4^{2(1)} = 9^2 - 4^2 = 81 - 16 = 65\). Is 65 divisible by 5? Yes, \(65 = 5 \times 13\). Is 65 divisible by 13? Yes, \(65 = 13 \times 5\). So for \(n=1\), the expression is divisible by both 5 and 13. Case 2: Let \(n=2\). \(9^{2(2)} - 4^{2(2)} = 9^4 - 4^4 = (9^2)^2 - (4^2)^2 = 81^2 - 16^2\). Using the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\): \(81^2 - 16^2 = (81-16)(81+16) = (65)(97)\). Since 65 is a factor, and 65 is divisible by both 5 and 13, the expression \((65)(97)\) is also divisible by both 5 and 13. Step 3: General Proof As shown in Step 1, the expression \(9^{2n} - 4^{2n}\) can be written as \(81^n - 16^n\). We know that for any positive integer \(n\), \(a^n - b^n\) is always divisible by \(a-b\). Here \(a=81\) and \(b=16\). So \(a-b = 81-16 = 65\). Since 65 is divisible by 5 (as \(65 = 5 \times 13\)) and divisible by 13 (as \(65 = 13 \times 5\)), it follows that \(9^{2n} - 4^{2n}\) is always divisible by both 5 and 13.