What is the remainder when 2 1000000 is divided by 7?
2
Finding the Remainder Using Modular Arithmetic
The problem asks for the remainder when \(2^{1000000}\) is divided by 7. This type of problem can be solved efficiently using modular arithmetic, specifically by looking for a pattern in the powers of the base number when divided by the divisor.
Step-by-Step Solution to Find the Remainder
We want to find \(2^{1000000} \pmod{7}\). Let's calculate the first few powers of 2 modulo 7:
- \(2^1 \equiv 2 \pmod{7}\)
- \(2^2 \equiv 4 \pmod{7}\)
- \(2^3 \equiv 8 \equiv 1 \pmod{7}\)
Now let's continue the pattern using the fact that \(2^3 \equiv 1 \pmod{7}\):
- \(2^4 \equiv 2^3 \cdot 2^1 \equiv 1 \cdot 2 \equiv 2 \pmod{7}\)
- \(2^5 \equiv 2^3 \cdot 2^2 \equiv 1 \cdot 4 \equiv 4 \pmod{7}\)
- \(2^6 \equiv 2^3 \cdot 2^3 \equiv 1 \cdot 1 \equiv 1 \pmod{7}\)
We can see that the remainders follow a repeating cycle of 2, 4, 1. The cycle length is 3, because \(2^3 \equiv 1 \pmod{7}\).
Using the Cycle Length to Find the Remainder
Since the pattern of remainders repeats every 3 powers, we can determine the remainder for a very large exponent like 1000000 by looking at the remainder of the exponent when divided by the cycle length (which is 3). We need to calculate \(1000000 \pmod{3}\).
To find the remainder of 1000000 divided by 3, we can use the divisibility rule for 3: a number is divisible by 3 if the sum of its digits is divisible by 3. The sum of the digits of 1000000 is \(1+0+0+0+0+0+0 = 1\).
Since \(1 \pmod{3}\) is 1, we have \(1000000 \equiv 1 \pmod{3}\).
Alternatively, we can note that \(10 \equiv 1 \pmod{3}\). Thus, \(10^{k} \equiv 1^k \equiv 1 \pmod{3}\) for any positive integer \(k\). So, \(1000000 = 10^6 \equiv 1 \pmod{3}\).
Since \(1000000 \equiv 1 \pmod{3}\), this means \(2^{1000000} \pmod{7}\) will behave like \(2^1 \pmod{7}\).
Looking back at our first few calculations:
- \(2^1 \pmod{7} \equiv 2\)
- \(2^2 \pmod{7} \equiv 4\)
- \(2^3 \pmod{7} \equiv 1\)
The exponent 1000000 corresponds to the first position in the cycle (because \(1000000 \equiv 1 \pmod{3}\)). The remainder at the first position is 2.
Therefore, \(2^{1000000} \equiv 2 \pmod{7}\).
Conclusion
The remainder when \(2^{1000000}\) is divided by 7 is 2.
| Exponent (\(n\)) | \(2^n\) | \(2^n \pmod{7}\) |
|---|---|---|
| 1 | 2 | 2 |
| 2 | 4 | 4 |
| 3 | 8 | 1 |
| 4 | 16 | 2 |
| 5 | 32 | 4 |
| 6 | 64 | 1 |
The pattern of remainders is 2, 4, 1, which repeats every 3 terms. Since the exponent 1000000 has a remainder of 1 when divided by 3 (\(1000000 = 3 \times 333333 + 1\)), the remainder of \(2^{1000000}\) when divided by 7 is the same as the remainder of \(2^1\) when divided by 7, which is 2.
Revision Table: Modular Arithmetic Concepts
| Concept | Explanation | Example |
|---|---|---|
| Modular Arithmetic (\(a \equiv b \pmod{m}\)) | \(a\) is congruent to \(b\) modulo \(m\) if \(a\) and \(b\) have the same remainder when divided by \(m\). Equivalently, \(m\) divides \(a-b\). | \(8 \equiv 1 \pmod{7}\) because \(8 \div 7\) has remainder 1, and \(1 \div 7\) has remainder 1. Also, \(7\) divides \(8-1=7\). |
| Properties of Congruence | If \(a \equiv b \pmod{m}\) and \(c \equiv d \pmod{m}\), then:
|
Since \(2^3 \equiv 1 \pmod{7}\), then \((2^3)^k \equiv 1^k \equiv 1 \pmod{7}\) for any \(k\). |
| Finding Cycle/Period | For \(a^n \pmod{m}\), the remainders often repeat in a cycle. The smallest positive integer \(k\) such that \(a^k \equiv 1 \pmod{m}\) is the order of \(a\) modulo \(m\) (if it exists and gcd(a,m)=1). This \(k\) is the length of the repeating cycle starting from \(a^1\). | For \(2^n \pmod{7}\), the cycle is 2, 4, 1. The length is 3, and \(2^3 \equiv 1 \pmod{7}\). |
Additional Information: Fermat's Little Theorem
For a prime number \(p\) and any integer \(a\) not divisible by \(p\), Fermat's Little Theorem states that \(a^{p-1} \equiv 1 \pmod{p}\).
In this problem, the modulus is 7, which is a prime number. The base is 2, which is not divisible by 7. According to Fermat's Little Theorem, \(2^{7-1} \equiv 2^6 \equiv 1 \pmod{7}\).
This confirms our earlier finding that the cycle length for powers of 2 modulo 7 related to 1 is 6 (or a divisor of 6). We found \(2^3 \equiv 1 \pmod{7}\), which is a smaller cycle length, 3. The cycle is indeed 2, 4, 1, which has length 3.
We needed to calculate \(2^{1000000} \pmod{7}\). We wrote \(1000000 = 3 \times 333333 + 1\).
So, \(2^{1000000} = 2^{(3 \times 333333 + 1)} = 2^{3 \times 333333} \cdot 2^1 = (2^3)^{333333} \cdot 2^1\).
Taking this modulo 7:
\(2^{1000000} \equiv (2^3)^{333333} \cdot 2^1 \pmod{7}\)
Since \(2^3 \equiv 1 \pmod{7}\), we substitute this in:
\(2^{1000000} \equiv (1)^{333333} \cdot 2 \pmod{7}\)
\(2^{1000000} \equiv 1 \cdot 2 \pmod{7}\)
\(2^{1000000} \equiv 2 \pmod{7}\)
This method using the remainder of the exponent \(1000000 \pmod{3}\) is a direct application of the property \(a^{kc+r} \equiv (a^c)^k \cdot a^r \equiv 1^k \cdot a^r \equiv a^r \pmod m\) where \(c\) is the cycle length such that \(a^c \equiv 1 \pmod m\) and \(k\) is the quotient and \(r\) is the remainder when the exponent is divided by \(c\).
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