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Chapter 6: Problem 475

For each \(n \in N, 2^{3 n}-7 n-1\) is divisible by (a) 64 (b) 36 (c) 49 (d) 25

Short Answer

Expert verified
After a step-by-step analysis, it can be concluded that the given expression \(2^{3n} - 7n - 1\) is not consistently divisible by any of the given options (a) 64, (b) 36, (c) 49, or (d) 25 for all natural numbers \(n\).

Step by step solution

01

Observe powers of 2

Notice that for every natural number \(n\), the term \(2^{3n}\) will always be a power of 2 (an even number). Therefore, the difference \(2^{3n} - 7n\) will also be an even number. Subtracting 1 from an even number gives an odd number. Hence, \(2^{3n} - 7n - 1\) will always be odd.
02

Eliminate options that are even

Since \(2^{3n} - 7n - 1\) is always odd, it cannot be divisible by any even number. In the given options, 64 and 36 are even numbers. Therefore, they can be eliminated. We are now left with two options: (c) 49 and (d) 25.
03

Test for smaller values of n

We can test the divisibility for smaller values of \(n\) for the remaining options. However, we should first try \(n=1\) and see which options satisfy the condition. \(2^{3(1)} - 7(1) - 1 = 8 - 7 - 1 = 0\) This shows that the expression is not divisible by either 49 or 25 when \(n = 1\). Now, let's try \(n = 2\). \(2^{3(2)} - 7(2) - 1 = 64 - 14 - 1 = 49\) This shows that the expression is divisible by 49 but not 25 for \(n = 2\). Therefore, we can conclude that the expression is divisible by 49 for some values of \(n\). However, we should check more cases before deciding which option is correct.
04

Analyze the divisibility by 49 for larger values of n

We can rewrite the given expression as follows: \(2^{3n} - 7n - 1 = 2^{3n} - (32n + 25n) - 1 = 2^{3n} - 32n - 1 - 25n\) Now, observe that when \(n\) is odd, \(2^{3n}\) is a power of 2, and hence, an even number. Furthermore, the term \(32n\) will also be even, as 32 is even. However, the term \(25n\) will be odd since 25 is odd. Therefore, when we add the terms, their sum will be odd. Subtracting 1 from an odd number gives an even number. Since 49 is odd, the expression cannot be divisible by 49 when \(n\) is odd. Therefore, we can conclude that there is no consistent divisibility by either 25 or 49.
05

Conclusion

After step-by-step analysis, we can conclude that there is no consistent divisibility for the given options. Therefore, the given expression is not divisible by any of the given options (a) 64, (b) 36, (c) 49, or (d) 25 for all natural numbers \(n\).

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