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Chapter 4: Problem 8

Let \(n \in \mathbf{Z}^{+}\)where \(n\) is odd and \(n\) is not divisible by 5 . Prove that there is a power of \(n\) whose units digit is 1 .

Short Answer

Expert verified
The powers of 1, 3, 7, 9 cetainly cycle to 1 and these are the only possibilities for the unit digit of an odd positive integer not divisible by 5. Thus, for an odd positive integer \(n\) not divisible by 5, there will always exist a power of \(n\) where the units digit is 1.

Step by step solution

01

Understanding the problem

This problem requires understanding both the properties of odd numbers and powers. The key insight for this problem is that the unit digit of a number's powers has a cyclical pattern.
02

Recognizing the unit digit cycle

For any given number, the unit digit of its powers cycle every 4 terms. This is because for any unit digit \(n\), \(n^4\) ends in \(n\). Since \(n\) is not divisible by 5, it cannot end in 0 or 5. Therefore, \(n\) must end in either 1, 3, 7, or 9. Given that \(n\) is an odd number, it cannot end in an even number. Therefore, the only possible candidates for the unit digit of \(n\) can be 1, 3, 7, or 9. The powers of these numbers cycle as follows: \n\n For \(n = 1, 1^1 = 1, 1^2 = 1, 1^3 = 1, 1^4 = 1, \ldots\):\n\n For \(n = 3, 3^1 = 3, 3^2 = 9, 3^4 = 1, 3^4 = 7, \ldots\):\n\n For \(n = 7, 7^1 = 7, 7^2 = 9, 7^3 = 3, 7^4 = 1, \ldots\):\n\n And for \(n = 9, 9^1 = 9, 9^2 = 1, 9^3 = 9, 9^4 = 1, \ldots\).
03

Conclusion

Given the above analysis, we can assert that for any \(n\) fulfilling the given criteria, i.e., \(n\) is an odd positive integer, the units digit of \(n\) is not 0 or 5 and will eventually cycle to 1 no matter the unit digit of \(n\) originally. Hence, there always exists a power of \(n\) such that the unit digit of that power is 1.

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Most popular questions from this chapter

If \(p\) is any prime, prove that \(\sqrt[3]{p}\) is irrational.

For \(n \in \mathbf{Z}^{+}\), prove each of the following by mathematical induction: a) \(5 \mid\left(n^{5}-n\right)\) b) \(6 \mid\left(n^{3}+5 n\right)\)

If \(a, b, c \in \mathbf{Z}^{+}\)and \(a \mid b c\), does it follow that \(a \mid b\) or \(a \mid c\) ?

Determine the smallest perfect square that is divisible by \(7 !\)

For any \(x \in \mathbf{R},|x|=\sqrt{x^{2}}=\left\\{\begin{array}{r}x, \text { if } x \geq 0 \\ -x, \text { if } x<0\end{array}\right\\}\), and \(-|x| \leq x \leq|x|\). Consequently, \(|x+y|^{2}=\) \((x+y)^{2}=x^{2}+2 x y+y^{2} \leq x^{2}+2|x||y|+y^{2}=|x|^{2}+2|x||y|+|y|^{2}=(|x|+|y|)^{2}\), and \(|x+y|^{2} \leq(|x|+|y|)^{2} \Rightarrow|x+y| \leq|x|+|y|\), for any \(x, y \in \mathbf{R}\). Prove that if \(n \in \mathbf{Z}^{+}, n \geq 2\), and \(x_{1}, x_{2}, \ldots, x_{n} \in \mathbf{R}\), then $$ \left|x_{1}+x_{2}+\cdots+x_{n}\right| \leq\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right| . $$
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