When every even power of every odd integer (greater than 1) is divided by 8, what is the remainder ?
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Finding the Remainder: Odd Integer Even Power Divided by 8
This question asks for the remainder when we divide an even power of any odd integer (that is greater than 1) by 8. Let's break down the problem and find a general solution.
Understanding the Problem
We are dealing with:
- Odd integers greater than 1. Examples are 3, 5, 7, 9, 11, and so on.
- Even powers. Examples are the power of 2, 4, 6, 8, and so on.
- The division is by the number 8.
- We need to find the remainder of this division.
Let's denote an odd integer greater than 1 as \(n\) and an even power as \(2k\), where \(k\) is a positive integer (\(k \ge 1\)). We want to find the remainder of \(n^{2k}\) when divided by 8.
Exploring Examples
Let's test a few cases with specific odd integers greater than 1 and even powers.
- Case 1: Odd integer \(n=3\), Even power \(2k=2\) (i.e., \(k=1\)). \(n^{2k} = 3^2 = 9\). When 9 is divided by 8, the remainder is 1 (\(9 = 1 \times 8 + 1\)).
- Case 2: Odd integer \(n=3\), Even power \(2k=4\) (i.e., \(k=2\)). \(n^{2k} = 3^4 = 81\). When 81 is divided by 8, the remainder is 1 (\(81 = 10 \times 8 + 1\)).
- Case 3: Odd integer \(n=5\), Even power \(2k=2\) (i.e., \(k=1\)). \(n^{2k} = 5^2 = 25\). When 25 is divided by 8, the remainder is 1 (\(25 = 3 \times 8 + 1\)).
- Case 4: Odd integer \(n=7\), Even power \(2k=2\) (i.e., \(k=1\)). \(n^{2k} = 7^2 = 49\). When 49 is divided by 8, the remainder is 1 (\(49 = 6 \times 8 + 1\)).
In all these examples, the remainder is 1. This suggests that the remainder might always be 1.
General Proof
Let's prove this mathematically for any odd integer \(n > 1\) and any even power \(2k\), where \(k \ge 1\).
Any odd integer \(n\) can be written in the form \(n = 2m + 1\) for some integer \(m\). Since \(n > 1\), \(2m + 1 > 1\), which means \(2m > 0\), so \(m\) must be a positive integer (\(m \ge 1\)).
Let's first consider the square of an odd integer, \(n^2\):
\(n^2 = (2m + 1)^2\)
Expanding this, we get:
\(n^2 = (2m)^2 + 2(2m)(1) + 1^2\)
\(n^2 = 4m^2 + 4m + 1\)
\(n^2 = 4m(m + 1) + 1\)
Now, consider the term \(m(m + 1)\). This is the product of two consecutive integers (\(m\) and \(m+1\)). The product of any two consecutive integers is always even. Therefore, \(m(m + 1)\) can be written as \(2p\) for some integer \(p\). Since \(m \ge 1\), \(m+1 \ge 2\), so \(m(m+1) \ge 2\), meaning \(p\) is a positive integer (\(p \ge 1\)).
Substitute \(m(m+1) = 2p\) back into the expression for \(n^2\):
\(n^2 = 4(2p) + 1\)
\(n^2 = 8p + 1\)
This form \(8p + 1\) shows that when the square of any odd integer (\(n > 1\)) is divided by 8, the remainder is always 1.
Now, we need to consider any even power \(n^{2k}\), where \(2k\) is an even power (\(k \ge 1\)). We can write \(n^{2k}\) as \((n^2)^k\).
Since \(n^2 = 8p + 1\), we have:
\(n^{2k} = (8p + 1)^k\)
Let's expand \((8p + 1)^k\) using the binomial theorem or simply consider its structure:
Any term in the expansion of \((a+b)^k\) will be of the form \(\binom{k}{j} a^j b^{k-j}\). In our case, \(a=8p\) and \(b=1\).
\((8p + 1)^k = \binom{k}{0}(8p)^0(1)^k + \binom{k}{1}(8p)^1(1)^{k-1} + \binom{k}{2}(8p)^2(1)^{k-2} + \dots + \binom{k}{k}(8p)^k(1)^0\)
\((8p + 1)^k = 1 + k(8p) + \frac{k(k-1)}{2}(64p^2) + \dots + (8p)^k\)
We can see that the first term is 1. Every subsequent term contains a factor of \(8p\) raised to a power of 1 or more. This means every term after the first one is a multiple of 8.
So, \((8p + 1)^k = 1 + (\text{a sum of terms, each divisible by 8})\)
\((8p + 1)^k = 1 + 8 \times (\text{some integer})\)
Thus, \((8p + 1)^k\) is of the form \(8Q + 1\) for some integer \(Q\). This shows that when \((8p + 1)^k\) is divided by 8, the remainder is always 1.
Since \(n^{2k} = (n^2)^k\) and \(n^2 \equiv 1 \pmod{8}\), it follows that \(n^{2k} \equiv 1^k \equiv 1 \pmod{8}\).
This holds true for any odd integer \(n > 1\) and any even power \(2k\) (\(k \ge 1\)).
The remainder is consistently 1.
Conclusion
When every even power of every odd integer (greater than 1) is divided by 8, the remainder is always 1.
Revision Table: Key Concepts
| Concept | Explanation |
|---|---|
| Odd Integer (\(n > 1\)) | An integer not divisible by 2, greater than 1 (e.g., 3, 5, 7...). Can be written as \(2m+1\) for \(m \ge 1\). |
| Even Power (\(2k\)) | A power that is an even number (e.g., \(n^2\), \(n^4\), \(n^6\)...). Can be written as \(2k\) for \(k \ge 1\). |
| Modular Arithmetic | A system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value (the modulus). \(a \equiv b \pmod{m}\) means \(a\) and \(b\) have the same remainder when divided by \(m\). |
| Square of Odd Integer | \((2m+1)^2 = 4m(m+1) + 1\). Since \(m(m+1)\) is even (\(=2p\)), the square is \(4(2p) + 1 = 8p+1\). |
| Even Power as \((n^2)^k\) | An even power \(n^{2k}\) can be seen as the \(k\)-th power of the square of the odd integer. |
| Remainder Property | If \(a \equiv 1 \pmod{m}\), then \(a^k \equiv 1^k \equiv 1 \pmod{m}\) for any positive integer \(k\). |
Additional Information: Properties of Odd and Even Numbers
Understanding the properties of odd and even numbers is fundamental in number theory problems like this one. Here are some key points:
- An even number can be written as \(2m\) for some integer \(m\).
- An odd number can be written as \(2m+1\) for some integer \(m\).
- The product of two consecutive integers is always even. \(m \times (m+1) = \text{even}\). This is because either \(m\) is even or \(m+1\) is even, and an even number times any integer is even.
- The square of an odd number is always odd. \((2m+1)^2 = 4m^2 + 4m + 1 = 2(2m^2 + 2m) + 1\).
- Any power of an odd number is always odd.
- The property \(n^2 \equiv 1 \pmod{8}\) for any odd integer \(n\) is a very useful result. It means that when you square any odd number (like 1, 3, 5, 7, 9...), the remainder when divided by 8 is always 1.
- \(1^2 = 1 \equiv 1 \pmod{8}\)
- \(3^2 = 9 = 1 \times 8 + 1 \equiv 1 \pmod{8}\)
- \(5^2 = 25 = 3 \times 8 + 1 \equiv 1 \pmod{8}\)
- \(7^2 = 49 = 6 \times 8 + 1 \equiv 1 \pmod{8}\)
- \(9^2 = 81 = 10 \times 8 + 1 \equiv 1 \pmod{8}\)
- Once we know \(n^2\) has a remainder of 1 when divided by 8, any higher even power \(n^{2k} = (n^2)^k\) will also have a remainder of 1 when divided by 8 because \(1^k = 1\).
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