What is the remainder after dividing the number 37 1000 by 9 ?
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Finding the Remainder using Modular Arithmetic
The question asks us to find the remainder when the large number 37 raised to the power of 1000, written as \(37^{1000}\), is divided by 9. This type of problem can be efficiently solved using the concept of modular arithmetic.
Understanding Modular Arithmetic for Remainders
Modular arithmetic deals with remainders after division. The notation \(a \equiv b \pmod{m}\) means that when \(a\) is divided by \(m\), the remainder is the same as when \(b\) is divided by \(m\). It also implies that \(a-b\) is a multiple of \(m\).
To find the remainder of a power like \(a^k\) when divided by \(m\), we can often simplify the base \(a\) first with respect to the modulus \(m\).
Step-by-Step Solution for \(37^{1000} \pmod{9}\)
Let's break down the problem:
- Simplify the base modulo the divisor: We first find the remainder when the base, 37, is divided by the divisor, 9.
- Divide 37 by 9: \(37 \div 9\).
- \(9 \times 4 = 36\).
- The remainder is \(37 - 36 = 1\).
- In modular notation, this is \(37 \equiv 1 \pmod{9}\). This means 37 and 1 have the same remainder when divided by 9.
- Apply the property of modular exponentiation: A key property in modular arithmetic states that if \(a \equiv b \pmod{m}\), then \(a^k \equiv b^k \pmod{m}\) for any positive integer \(k\).
- Since we know \(37 \equiv 1 \pmod{9}\), we can raise both sides of the congruence to the power of 1000.
- So, \(37^{1000} \equiv 1^{1000} \pmod{9}\).
- Calculate the simplified power modulo the divisor: Now we need to calculate \(1^{1000} \pmod{9}\).
- Any positive integer power of 1 is always 1. \(1^{1000} = 1\).
- So, \(1^{1000} \equiv 1 \pmod{9}\).
- State the final remainder: Combining the results from the previous steps, we have:
- \(37^{1000} \equiv 1^{1000} \pmod{9}\)
- \(37^{1000} \equiv 1 \pmod{9}\)
- This means when \(37^{1000}\) is divided by 9, the remainder is 1.
Final Answer Derivation
By simplifying the base modulo 9 (\(37 \equiv 1 \pmod{9}\)), we found that calculating the remainder of \(37^{1000}\) divided by 9 is equivalent to calculating the remainder of \(1^{1000}\) divided by 9. Since \(1^{1000}\) is simply 1, the remainder is 1.
Therefore, the remainder after dividing the number \(37^{1000}\) by 9 is 1.
| Expression | Modular Congruence | Explanation |
|---|---|---|
| \(37 \div 9\) | \(37 \equiv 1 \pmod{9}\) | 37 leaves a remainder of 1 when divided by 9. |
| \(37^{1000} \pmod{9}\) | \(37^{1000} \equiv 1^{1000} \pmod{9}\) | Using the property \(a \equiv b \implies a^k \equiv b^k\). |
| \(1^{1000}\) | \(1^{1000} = 1\) | Any power of 1 is 1. |
| \(37^{1000} \pmod{9}\) | \(37^{1000} \equiv 1 \pmod{9}\) | The final remainder is 1. |
Revision Table: Key Concepts in Modular Arithmetic
| Concept | Description | Notation |
|---|---|---|
| Modulo Operation | Finding the remainder after division of one number by another. | \(a \pmod{m}\) (read as "a modulo m") |
| Congruence Modulo m | Two integers \(a\) and \(b\) are congruent modulo \(m\) if they have the same remainder when divided by \(m\). Equivalently, if \(a-b\) is a multiple of \(m\). | \(a \equiv b \pmod{m}\) |
| Properties of Congruence | If \(a \equiv b \pmod{m}\) and \(c \equiv d \pmod{m}\):
|
Additional Information: Modular Exponentiation
Modular exponentiation is a fundamental operation in number theory and is used in various fields like cryptography. The problem \(a^b \pmod{m}\) can be computed efficiently, especially when \(b\) is very large, by repeatedly using the property \((x \times y) \pmod{m} = ((x \pmod{m}) \times (y \pmod{m})) \pmod{m}\). In our specific case, the base \(37\) simplified to \(1 \pmod{9}\), which made the calculation very simple. For more complex bases and moduli, one might use techniques like the method of repeated squaring or properties derived from Euler's Totient Theorem or Fermat's Little Theorem, although these advanced concepts were not necessary for this particular problem.
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