Understanding the Problem: Sum of Factorials and Remainder

The problem asks for the remainder when a large sum, \(S = 0! + 1! + 2! + 3! + 4! + \dots + 100!\), is divided by 60. To solve this, we need to understand the concept of factorials and how they behave under modular arithmetic, specifically modulo 60.

Calculating Initial Factorial Values

Let's calculate the value of the first few terms in the sum S:

• \(0! = 1\)
• \(1! = 1\)
• \(2! = 2 \times 1 = 2\)
• \(3! = 3 \times 2 \times 1 = 6\)
• \(4! = 4 \times 3 \times 2 \times 1 = 24\)
• \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
• \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\)

Analyzing Factorials Modulo 60

We are interested in the remainder when S is divided by 60. This means we need to find S modulo 60. We can find the remainder of each term \(n!\) when divided by 60 and then sum these remainders modulo 60.

Let's find the remainders for the first few factorials:

• \(0! \pmod{60} = 1 \pmod{60} = 1\)
• \(1! \pmod{60} = 1 \pmod{60} = 1\)
• \(2! \pmod{60} = 2 \pmod{60} = 2\)
• \(3! \pmod{60} = 6 \pmod{60} = 6\)
• \(4! \pmod{60} = 24 \pmod{60} = 24\)

Now consider \(5!\):

• \(5! = 120\)
• To find \(5! \pmod{60}\), we divide 120 by 60. \(120 = 2 \times 60\).
• So, \(5! \pmod{60} = 0\).

What about \(6!\)?

• \(6! = 6 \times 5! = 6 \times 120 = 720\)
• To find \(6! \pmod{60}\), we divide 720 by 60. \(720 = 12 \times 60\).
• So, \(6! \pmod{60} = 0\).

Identifying the Pattern for Factorials Modulo 60

Notice that \(60 = 5 \times 12\). Also, \(60 = 5 \times 4 \times 3\). For a number to be divisible by 60, it must be divisible by its prime factors \(2^2\), 3, and 5. The number 60 can also be seen as the product of relatively prime numbers like 3, 4, and 5.

Let's look at the prime factors present in \(n!\) as n increases:

• \(1!\) has no prime factors greater than 1.
• \(2!\) has a factor of 2.
• \(3!\) has factors of 2 and 3.
• \(4!\) has factors of \(2^3\) and 3.
• \(5!\) has factors of \(2^3\), 3, and 5.

The number 60 has prime factors \(2^2\), 3, and 5. We see that \(4!\) has \(2^3\) and 3, but not 5, so it's not divisible by 60. However, \(5!\) contains \(2^3\), 3, and 5 as factors, which is more than enough to be divisible by \(2^2\), 3, and 5. Thus, \(5!\) is divisible by 60.

For any integer \(n\) greater than or equal to 5, \(n!\) can be written as \(n \times (n-1) \times \dots \times 5!\). Since \(5!\) is divisible by 60, any multiple of \(5!\) will also be divisible by 60. Therefore, for all \(n \ge 5\), \(n!\) is divisible by 60, which means \(n! \pmod{60} = 0\).

Calculating the Remainder of the Sum S

The sum is \(S = 0! + 1! + 2! + 3! + 4! + 5! + \dots + 100!\).

When finding the remainder of S divided by 60, we can sum the remainders of each term when divided by 60:

Using the remainders we found:

We only need to sum the remainders for terms up to 4!:

The remainder when S is divided by 60 is 34.

Summary of Calculation

We calculated the remainders of the first few factorials when divided by 60:

For \(n \ge 5\), \(n! \pmod{60} = 0\).

The sum modulo 60 is the sum of the first few remainders:

Final Answer

The remainder when the sum S is divided by 60 is 34.

Revision Table: Sum of Factorials Remainder

This table helps summarize the key steps and values involved in finding the remainder of the sum of factorials when divided by 60.

Additional Information: Modular Arithmetic and Factorials

Modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value, called the modulus. When we say "a modulo m", we mean the remainder when a is divided by m.

Factorials grow very quickly. \(n!\) is the product of all positive integers up to n. The properties of factorials related to divisibility are important in number theory problems. If a number k is a factor of m, and n is large enough such that \(n!\) contains all prime factors of k with at least the same power as in k, then \(n!\) will be divisible by k.

In this problem, 60 = \(2^2 \times 3 \times 5\). For \(n!\) to be divisible by 60, it must contain factors \(2^2\), 3, and 5. The first factorial that contains a factor of 5 is 5!. Since 5! also contains \(2^2\) (from 4!) and 3, 5! is divisible by 60. Any subsequent factorial, \(n!\) for \(n > 5\), will include 5! as a factor, and thus will also be divisible by 60.