To solve the expression \((1! + 2! + 3! + 4! + 5! + 6!) \mod 7\), we'll break it down step by step.

1. Calculate each factorial:
   - \(1! = 1\)
   - \(2! = 2\)
   - \(3! = 6\)
   - \(4! = 24\)
   - \(5! = 120\)
   - \(6! = 720\)

2. Add these factorials together:
   \[
   1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873
   \]

3. Find \(873 \mod 7\):
   - First, divide 873 by 7 to find the remainder:
   \[
   873 \div 7 = 124 \text{ with a remainder of } 5
   \]

So, \((1! + 2! + 3! + 4! + 5! + 6!) \mod 7 = 5\).

Would you like more details on any part of this solution? Here are some related questions you might find interesting:

1. What is the general formula for the factorial of a number \(n\)?
2. How do modular arithmetic rules apply to factorials?
3. Can you find the modular of other sums like \((1! + 2! + 3! + 4! + 5!) \mod 6\)?
4. How would the answer change if we included \(7!\) in the sum?
5. What is the significance of modular arithmetic in number theory?
6. How can the properties of factorials be used in combinatorics?
7. What are some common applications of factorials in probability?
8. How does modular arithmetic help in solving congruences?

**Tip:** Modular arithmetic often simplifies calculations in number theory, especially when dealing with large numbers like factorials.

Learn how to solve the expression (1! + 2! + 3! + 4! + 5! + 6!) mod 7 step-by-step with detailed factorial calculations and modular arithmetic. This problem explores factorials and modular arithmetic rules, suitable for advanced high school students.

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