Question

If 'n!' denotes the product of the integers 1 through 'n', what is the remainder when (1! + 2! + 3! + 4! + 5! + 6! + ....) is divided by 9?

• A. 8
• B. 1
• C. 2
• D. 0

Hint:

Check where the terms start becoming divisible by the divisor to simplify.

Answer 1

The Correct Option is D

Step 1: Concept
Factorials and divisibility.

Step 2: Analysis
$6! = 720$, which is divisible by 9. Any $n!$ for $n \geq 6$ is divisible by 9 because it contains $3 \times 6 = 18$ as a factor.

Step 3: Calculation
Sum modulo 9: $(1! + 2! + 3! + 4! + 5!) \pmod 9$.
$(1 + 2 + 6 + 24 + 120) = 153$.
$153 \div 9 = 17$ with remainder 0.

Step 4: Conclusion
The remainder is 0. Final Answer: (D)