To find the remainder when the product $2005 \cdot 2006 \cdot 2007 \cdot 2008 \cdot 2009 \cdot 2010$ is divided by 11, we will apply Fermat's Little Theorem and properties of congruences.

Step 1: Find each term modulo 11

First, calculate each number modulo 11: 2005 \equiv 3 \pmod{11} \quad \text{(since 2005 - 3 = 2002 is divisible by 11)} 2006 \equiv 4 \pmod{11} \quad \text{(since 2006 - 4 = 2002)} 2007 \equiv 5 \pmod{11} \quad \text{(since 2007 - 5 = 2002)} 2008 \equiv 6 \pmod{11} \quad \text{(since 2008 - 6 = 2002)} 2009 \equiv 7 \pmod{11} \quad \text{(since 2009 - 7 = 2002)} 2010 \equiv 8 \pmod{11} \quad \text{(since 2010 - 8 = 2002)}

Step 2: Calculate the product modulo 11

We now need to find the product: $3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \pmod{11}$ We'll multiply step-by-step: $3 \cdot 4 = 12 \equiv 1 \pmod{11}$ $1 \cdot 5 = 5 \pmod{11}$ $5 \cdot 6 = 30 \equiv 8 \pmod{11}$ $8 \cdot 7 = 56 \equiv 1 \pmod{11}$ $1 \cdot 8 = 8 \pmod{11}$

Final Answer

The remainder when $2005 \cdot 2006 \cdot 2007 \cdot 2008 \cdot 2009 \cdot 2010$ is divided by 11 is: $\boxed{8}$

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Related Questions

1. What is Fermat's Little Theorem, and how is it applied in modular arithmetic problems?
2. How can we generalize finding remainders of large products modulo smaller numbers?
3. Can this method be used for any prime modulus, or are there exceptions?
4. How does modular arithmetic simplify calculations in number theory?
5. Why is understanding modular arithmetic crucial in cryptography?

Tip

When dealing with products and sums modulo a number, breaking down complex expressions step-by-step simplifies calculations significantly.