Let's find the remainder when the sum 1! + 3! + 5! + 7! + ... + 199! is divided by 24. To do this, we can find the remainder of each term when divided by 24 and then sum the remainders.

Understanding Factorials and Remainders

A factorial, denoted by n!, is the product of all positive integers up to n. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120.

The remainder is the amount left over after division. When we divide a number 'a' by 'b', the remainder 'r' satisfies a = qb + r, where q is the quotient and 0 ≤ r < b. We write this as a ≡ r (mod b).

Calculating Remainders of First Few Terms

Let's calculate the factorials for the first few odd numbers and their remainders when divided by 24.

Identifying the Pattern in Remainders

We observe that:

• 1! mod 24 = 1
• 3! mod 24 = 6
• 5! mod 24 = 0
• 7! mod 24 = 0

Let's consider the terms from 5! onwards. Any factorial n! for n ≥ 4 contains both 4 and 3 as factors. Since 4 and 3 are factors, their product 4 × 3 = 12 is a factor of n!. Also, for n ≥ 4, n! is an even number. Therefore, for n ≥ 4, n! is divisible by 2 × 12 = 24.

In the given sum, 1! + 3! + 5! + 7! + ... + 199!, the terms 5!, 7!, 9!, ..., all the way up to 199!, are factorials of numbers greater than or equal to 5. Since 5 > 4, all these terms (5!, 7!, 9!, ..., 199!) are divisible by 24. This means their remainder when divided by 24 is 0.

So, for any odd number k where k ≥ 5, we have k! ≡ 0 (mod 24).

Calculating the Remainder of the Sum

The sum is S = 1! + 3! + 5! + 7! + ... + 199!.

The remainder of the sum when divided by 24 is the sum of the remainders of each term when divided by 24, taken modulo 24.

S mod 24 = (1! mod 24 + 3! mod 24 + 5! mod 24 + 7! mod 24 + ... + 199! mod 24) mod 24

Using the remainders we found:

S mod 24 = (1 + 6 + 0 + 0 + ... + 0) mod 24

S mod 24 = (1 + 6) mod 24

S mod 24 = 7 mod 24

The remainder is 7.

Thus, when 1! + 3! + 5! + 7! + ... + 199! is divided by 24, the remainder is 7.

Revision Table: Factorials and Remainders

Additional Information: Modular Arithmetic Properties

This problem uses concepts from modular arithmetic. Key properties include:

• Sum Property: (a + b) mod m ≡ ((a mod m) + (b mod m)) mod m
• Product Property: (a × b) mod m ≡ ((a mod m) × (b mod m)) mod m
• Divisibility: If 'a' is divisible by 'm', then a mod m = 0.

In this problem, we applied the sum property to find the remainder of the entire sum by adding the remainders of individual terms.

Understanding when a factorial n! is divisible by a number m is crucial for solving such problems. Specifically, n! is divisible by any prime number p ≤ n and by any composite number whose prime factors are all ≤ n, considering their powers. Since 24 = 2³ × 3, a number needs to have factors of 2, 2, 2, and 3 to be divisible by 24. For n ≥ 4, n! includes factors 1, 2, 3, 4, ..., n. This includes 3 and 4 (which contains 2²). The factor 2 is also present for n ≥ 2. So, for n ≥ 4, n! contains factors 2, 3, and 4, meaning it's divisible by 2 × 3 × 4 = 24. Thus, k! is divisible by 24 for any k ≥ 4.

Note that: \[ n! \text{ is divisible by } 24 \text{ for all } n \geq 4 \] So, 
All terms from \( 5! \) to \( 199! \) are divisible by 24 → contribute 0 to the remainder.

Only two terms are not divisible by 24: 
\( 1! = 1 \), \( 3! = 6 \) 
Sum = \( 1 + 6 = 7 \)

So, remainder = \( \boxed{7} \)

Answer:

7

Step 1: Understand Factorials Modulo 24

For any factorial 4! or higher: \[ n! \equiv 0 \ (\text{mod}\ 24) \quad \text{for}\ n \geq 4 \] because 24 = 4! and all higher factorials contain 4×3×2 as factors.

Step 2: Evaluate Relevant Terms

Only factorials below 4! contribute to the remainder: \[ 1! = 1 \\ 3! = 6 \] All other terms (5! to 199!) are ≡ 0 mod 24.

Step 3: Calculate the Sum Modulo 24

\[ 1! + 3! + 5! + \cdots + 199! \equiv 1 + 6 + 0 + \cdots + 0 \ (\text{mod}\ 24) \] \[ \equiv 7 \ (\text{mod}\ 24) \]

Final Answer:

The remainder is \[ \boxed{7} \].

Key Insights:

• 24 divides all factorials ≥4! completely
• Only 1! and 3! contribute to the remainder
• The pattern holds for any odd factorial series beyond 5!