If n is a positive integer, n! nd by multiplying the integers from 1 to n. For example, 4! = 4 × 3 × 2 × 1 = 24. What are the three rightmost digits of the sum 1! + 2!+ 3! + ·· · + 2020! ?

To solve for the three rightmost digits of the sum \(1! + 2! + 3! + \dots + 2020!\), we can use properties of factorials and modular arithmetic.

### Observations:
- **Factorials grow extremely fast**: Beyond \(10!\), factorials are very large numbers that end in multiple zeroes.
- The reason is that from \(10!\) onwards, each factorial includes at least one factor of \(10 = 2 \times 5\), so each term from \(10!\) onward ends in at least one zero. This means that for \(n \geq 10\), the factorial \(n!\) contributes nothing to the last three digits of the sum, as these will all be zero.

Thus, to determine the three rightmost digits of the sum \(1! + 2! + \dots + 2020!\), we only need to compute the sum of the factorials up to \(9!\), as all factorials from \(10!\) onwards will not affect the last three digits.

### Calculate the factorials up to \(9!\):
\[
\begin{aligned}
1! &= 1, \\
2! &= 2, \\
3! &= 6, \\
4! &= 24, \\
5! &= 120, \\
6! &= 720, \\
7! &= 5040, \\
8! &= 40320, \\
9! &= 362880.
\end{aligned}
\]

### Sum of factorials up to \(9!\):
Now, sum these values:
\[
1 + 2 + 6 + 24 + 120 + 720 + 5040 + 40320 + 362880 = 408113.
\]

### Last three digits:
We only need the last three digits of this sum:
\[
408113 \mod 1000 = 113.
\]

Thus, the three rightmost digits of the sum \(1! + 2! + 3! + \dots + 2020!\) are **113**.

Would you like any more details or further clarifications?

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Here are 5 related questions for further exploration:
1. What is the last non-zero digit of \(100!\)?
2. How many trailing zeroes does \(1000!\) have?
3. Find the sum of the digits of \(9!\).
4. What are the last two digits of \(50!\)?
5. What is the remainder when \(999!\) is divided by 10?

**Tip**: Factorials beyond \(10!\) typically result in many trailing zeroes because every multiple of \(5 \times 2\) produces a zero.

If n is a positive integer, n! is defined by multiplying the integers from 1 to n. For example, 4! = 4 × 3 × 2 × 1 = 24. What are the three rightmost digits of the sum 1! + 2!+ 3! + ·· · + 2020!?

Learn how to find the three rightmost digits of the sum 1! + 2! + ... + 2020! using properties of factorials and modular arithmetic. Explore why factorials beyond 10! contribute no significant digits and how the sum is computed up to 9!. Suitable for high school students in Grades 10-12.

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