It is given that 5 does not divide n - 1, n and n + 1, where n is a positive integer. Which one of the following is correct ?
5 divides (n2 + 1)
Understanding Divisibility When 5 Does Not Divide n-1, n, or n+1
The problem states that for a positive integer \(n\), the number 5 does not divide \(n-1\), \(n\), or \(n+1\). We need to find which of the given expressions is divisible by 5 under this condition.
Let's analyze the condition using modular arithmetic. When an integer \(n\) is divided by 5, the possible remainders are 0, 1, 2, 3, or 4. We can write this as \(n \pmod{5}\) can be 0, 1, 2, 3, or 4.
The condition "5 does not divide \(n\)" means \(n \pmod{5} \neq 0\).
The condition "5 does not divide \(n-1\)" means \(n-1 \pmod{5} \neq 0\). If \(n-1\) is divisible by 5, then \(n \equiv 1 \pmod{5}\). So, \(n \pmod{5} \neq 1\).
The condition "5 does not divide \(n+1\)" means \(n+1 \pmod{5} \neq 0\). If \(n+1\) is divisible by 5, then \(n \equiv -1 \equiv 4 \pmod{5}\). So, \(n \pmod{5} \neq 4\).
Combining these conditions, the remainder when \(n\) is divided by 5 cannot be 0, 1, or 4. Therefore, the possible remainders for \(n\) when divided by 5 are 2 or 3.
- Case 1: \(n \equiv 2 \pmod{5}\)
- Case 2: \(n \equiv 3 \pmod{5}\)
Now, let's evaluate each of the given options under these two cases using modular arithmetic to check for divisibility by 5 (i.e., checking if the result is \(\equiv 0 \pmod{5}\)).
Evaluating the Options
| Option | Expression | Case 1: \(n \equiv 2 \pmod{5}\) | Case 2: \(n \equiv 3 \pmod{5}\) | Divisible by 5? |
|---|---|---|---|---|
| 1 | \(n^2 + 1\) | \(n^2 \equiv 2^2 \equiv 4 \pmod{5}\). So, \(n^2 + 1 \equiv 4 + 1 \equiv 5 \equiv 0 \pmod{5}\). | \(n^2 \equiv 3^2 \equiv 9 \equiv 4 \pmod{5}\). So, \(n^2 + 1 \equiv 4 + 1 \equiv 5 \equiv 0 \pmod{5}\). | Yes (in both cases) |
| 2 | \(n^2 - 1\) | \(n^2 \equiv 4 \pmod{5}\). So, \(n^2 - 1 \equiv 4 - 1 \equiv 3 \pmod{5}\). | \(n^2 \equiv 4 \pmod{5}\). So, \(n^2 - 1 \equiv 4 - 1 \equiv 3 \pmod{5}\). | No (in both cases) |
| 3 | \(n^2 + n\) | \(n^2 \equiv 4 \pmod{5}\), \(n \equiv 2 \pmod{5}\). So, \(n^2 + n \equiv 4 + 2 \equiv 6 \equiv 1 \pmod{5}\). | \(n^2 \equiv 4 \pmod{5}\), \(n \equiv 3 \pmod{5}\). So, \(n^2 + n \equiv 4 + 3 \equiv 7 \equiv 2 \pmod{5}\). | No (in both cases) |
| 4 | \(n^2 - n\) | \(n^2 \equiv 4 \pmod{5}\), \(n \equiv 2 \pmod{5}\). So, \(n^2 - n \equiv 4 - 2 \equiv 2 \pmod{5}\). | \(n^2 \equiv 4 \pmod{5}\), \(n \equiv 3 \pmod{5}\). So, \(n^2 - n \equiv 4 - 3 \equiv 1 \pmod{5}\). | No (in both cases) |
From the table, we can see that only the expression \(n^2 + 1\) is divisible by 5 in both possible cases for \(n\). Therefore, 5 divides \((n^2 + 1)\).
Conclusion
Given that 5 does not divide \(n-1\), \(n\), or \(n+1\) for a positive integer \(n\), it must be true that 5 divides \((n^2 + 1)\).
Revision Table: Divisibility Concepts
| Concept | Description | Example |
|---|---|---|
| Divisibility | An integer \(a\) divides an integer \(b\) if there exists an integer \(k\) such that \(b = ak\). We write \(a | b\). | 3 divides 12 because \(12 = 3 \times 4\). |
| Modular Arithmetic | A system of arithmetic for integers, where numbers "wrap around" when they reach a certain value (the modulus). \(a \equiv b \pmod{m}\) means \(a - b\) is divisible by \(m\). | \(7 \equiv 2 \pmod{5}\) because \(7 - 2 = 5\), and 5 is divisible by 5. |
| Congruence Properties | If \(a \equiv b \pmod{m}\) and \(c \equiv d \pmod{m}\), then \(a+c \equiv b+d \pmod{m}\) and \(ac \equiv bd \pmod{m}\). | If \(n \equiv 2 \pmod{5}\), then \(n^2 \equiv 2^2 \equiv 4 \pmod{5}\). |
Additional Information: Working with Modulo 5
When working with modulo 5, any integer \(n\) can be represented by its remainder when divided by 5. This remainder is one of \{0, 1, 2, 3, 4\}.
- If \(n \equiv 0 \pmod{5}\), then 5 divides \(n\).
- If \(n \equiv 1 \pmod{5}\), then \(n-1 \equiv 1-1 \equiv 0 \pmod{5}\), so 5 divides \(n-1\).
- If \(n \equiv 2 \pmod{5}\), then \(n-1 \equiv 1 \pmod{5}\), \(n \equiv 2 \pmod{5}\), \(n+1 \equiv 3 \pmod{5}\). None of these are divisible by 5.
- If \(n \equiv 3 \pmod{5}\), then \(n-1 \equiv 2 \pmod{5}\), \(n \equiv 3 \pmod{5}\), \(n+1 \equiv 4 \pmod{5}\). None of these are divisible by 5.
- If \(n \equiv 4 \pmod{5}\), then \(n+1 \equiv 4+1 \equiv 5 \equiv 0 \pmod{5}\), so 5 divides \(n+1\).
The condition that 5 does not divide \(n-1\), \(n\), or \(n+1\) directly leads to \(n \pmod{5}\) being restricted to only 2 or 3. This is a key step in solving this type of divisibility problem.
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