What is the remainder when the sum 1 5+ 2 5+ 3 5+ 4 5+ 5 5is divided by 4 ?
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Calculating the Remainder of a Sum of Powers When Divided by 4
The problem asks for the remainder when the sum of the first five positive integers raised to the power of 5 is divided by 4. The sum is given by:
\[ \text{Sum} = 1^5 + 2^5 + 3^5 + 4^5 + 5^5 \]
To find the remainder when this sum is divided by 4, we can use the properties of modular arithmetic. Specifically, the remainder of a sum divided by a number is the same as the remainder of the sum of the individual remainders divided by that number.
Let's find the remainder for each term when divided by 4:
- For \(1^5\): \(1^5 = 1\). When 1 is divided by 4, the remainder is 1. We can write this as \(1^5 \equiv 1 \pmod{4}\).
- For \(2^5\): \(2^5 = 32\). When 32 is divided by 4, the remainder is 0, since \(32 = 4 \times 8\). We can write this as \(2^5 \equiv 0 \pmod{4}\).
- For \(3^5\): \(3^5 = 243\). To find the remainder when 243 is divided by 4, we can perform the division: \(243 = 4 \times 60 + 3\). The remainder is 3. Alternatively, we know \(3 \equiv -1 \pmod{4}\), so \(3^5 \equiv (-1)^5 \pmod{4}\). Since \((-1)^5 = -1\), we have \(3^5 \equiv -1 \pmod{4}\). In terms of positive remainders, \(-1 \equiv 3 \pmod{4}\). So, \(3^5 \equiv 3 \pmod{4}\).
- For \(4^5\): \(4^5 = 1024\). When 1024 is divided by 4, the remainder is 0, since \(1024\) is a multiple of 4 (\(1024 = 4 \times 256\)). We can write this as \(4^5 \equiv 0 \pmod{4}\). In general, any power of 4 (greater than or equal to 1) is divisible by 4.
- For \(5^5\): \(5^5 = 3125\). To find the remainder when 3125 is divided by 4, we can perform the division: \(3125 = 4 \times 781 + 1\). The remainder is 1. Alternatively, we know \(5 \equiv 1 \pmod{4}\), so \(5^5 \equiv 1^5 \pmod{4}\). Since \(1^5 = 1\), we have \(5^5 \equiv 1 \pmod{4}\).
Let's summarize the remainders for each term:
| Term | Value | Remainder when divided by 4 (\( \pmod{4} \)) |
|---|---|---|
| \(1^5\) | 1 | 1 |
| \(2^5\) | 32 | 0 |
| \(3^5\) | 243 | 3 |
| \(4^5\) | 1024 | 0 |
| \(5^5\) | 3125 | 1 |
Now, we find the remainder of the sum by summing the individual remainders and then finding the remainder of that sum when divided by 4.
Sum of remainders = \(1 + 0 + 3 + 0 + 1 = 5\).
Finally, we find the remainder when 5 is divided by 4.
\[ 5 \div 4 \]
We know that \(5 = 4 \times 1 + 1\). So, the remainder is 1.
Therefore, the remainder when the sum \(1^5 + 2^5 + 3^5 + 4^5 + 5^5\) is divided by 4 is 1.
Revision Table: Sum of Powers Remainder by 4
Here is a quick look at the key steps to find the remainder of the sum:
- Calculate the remainder of each term \(n^5\) when divided by 4.
- Sum these individual remainders.
- Find the remainder of this sum when divided by 4.
Additional Information: Modular Arithmetic and Remainders
Modular arithmetic is very useful for solving problems involving remainders. The expression \(a \equiv b \pmod{m}\) means that \(a\) and \(b\) have the same remainder when divided by \(m\). This is equivalent to saying that \(a - b\) is divisible by \(m\).
Properties used in this problem:
- If \(a \equiv b \pmod{m}\) and \(c \equiv d \pmod{m}\), then \(a+c \equiv b+d \pmod{m}\). This means we can add remainders.
- If \(a \equiv b \pmod{m}\), then \(a^k \equiv b^k \pmod{m}\) for any positive integer \(k\). This allows us to find the remainder of a power by finding the remainder of the base first. For example, since \(5 \equiv 1 \pmod{4}\), we have \(5^5 \equiv 1^5 \pmod{4}\).
- The remainder of a number \(N\) when divided by 4 can be found by looking at the remainder of the last two digits of \(N\) when divided by 4, or by checking its value modulo 4. For small numbers, direct calculation is easy. For larger numbers, using modular properties like \(n \pmod{4}\) for the base \(n\) simplifies finding \(n^5 \pmod{4}\).
Understanding these basic rules of modular arithmetic helps in solving problems efficiently without calculating large sums or powers directly.
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