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Chapter 6: Problem 459

For all \(n \geq 2, n^{2}\left(n^{4}-1\right)\) is divisible by (a) 60 (b) 50 (c) 40 (d) 70

Short Answer

Expert verified
The expression \(n^2(n^4 - 1)\) can be factored as \(n^2(n - 2)(n + 2)(n^2 + 4)\). Analyzing the factors, we find that the expression will always be divisible by 60 for all \(n \ge 2\). Thus, the answer is (a) 60.

Step by step solution

01

Factor the expression

First, we need to factor the expression \(n^{2}(n^{4}-1)\). We can rewrite the expression as: \[n^2(n^4-1) = n^2(n^4 - 2^4 + 2^4 - 1)\] Using the difference of squares formula, \((a^2 - b^2) = (a + b)(a - b)\), we can factor further: \[n^2(n^4 - 2^4 + 2^4 - 1) = n^2((n^4 - 2^4) + (2^4 - 1))\] \[= n^2(n^2 - 2^2)(n^2 + 2^2)\] \[= n^2(n - 2)(n + 2)(n^2 + 4)\]
02

Analyze factors

Now, we analyze each factor. - First factor, \(n^2\): Since \(n \ge 2\), \(n^2\) will have at least one factor of 2 and at least one factor of 4. Thus, the first factor is divisible by 4. - Second factor, \((n - 2)\): It has at least one factor of 3 because it is formed by subtracting 2 from a multiple of 3. - Third factor, \((n + 2)\): It has at least one factor of 5 because it is formed by adding 2 to a multiple of 5. - Fourth factor, \((n^2 + 4)\): This factor is a sum of even numbers so it is always even, meaning it has at least one factor of 2.
03

Determine the required multiple

We can now combine the factors of each factor to form the multiple: \[2^2 \times 3 \times 5 \times 2 = 60\] Since 60 contains all the factors we have identified, the expression will always be divisible by 60. Thus, the answer is: (a) 60

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