Find he units digit of {2013^1} + {2013^2} + {2013^3} + .... + {2013^{2013}}.

1

2

3

4

To find the units digit of the sum \(2013^1 + 2013^2 + 2013^3 + \dots + 2013^{2013}\), we first determine the units digit of \(2013^n\) for various powers \(n\).

Since only the last digit of the base affects the units digit, we only need to consider the units digit of 2013, which is 3. So, the problem reduces to finding the units digit of the sum \(3^1 + 3^2 + 3^3 + \dots + 3^{2013}\).

First, let's analyze the units digits of the powers of 3:

\[
\begin{aligned}
3^1 &\equiv 3 \pmod{10}, \\
3^2 &\equiv 9 \pmod{10}, \\
3^3 &\equiv 7 \pmod{10}, \\
3^4 &\equiv 1 \pmod{10}.
\end{aligned}
\]

Notice that the units digits repeat every four terms: \(3, 9, 7, 1\).

This sequence repeats with a cycle length of 4. Thus:

- \(3^{4k+1} \equiv 3 \pmod{10}\)
- \(3^{4k+2} \equiv 9 \pmod{10}\)
- \(3^{4k+3} \equiv 7 \pmod{10}\)
- \(3^{4k+4} \equiv 1 \pmod{10}\)

Now, to compute the sum of the units digits for all powers from 1 to 2013:

1. The number of complete cycles of 4 in the sequence from \(3^1\) to \(3^{2013}\) is \(\left\lfloor \frac{2013}{4} \right\rfloor = 503\) complete cycles.

2. Each complete cycle contributes:
   \[
   3 + 9 + 7 + 1 = 20
   \]
   to the sum of the units digits.

3. Therefore, the contribution from the complete cycles is:
   \[
   503 \times 20 = 10060
   \]

4. The remaining terms are \(3^{2013}\), since \(2013 \mod 4 = 1\), corresponding to the first term in the cycle, which has a units digit of 3.

Thus, the sum of the units digits of \(2013^1 + 2013^2 + \dots + 2013^{2013}\) has a units digit given by:

\[
10060 + 3 = 10063
\]

The units digit of 10063 is \(3\).

Therefore, the units digit of the sum \(2013^1 + 2013^2 + 2013^3 + \dots + 2013^{2013}\) is:

\[
\boxed{3}
\]

Would you like further details or have any questions? Here are 5 related questions to expand your understanding:

1. How do you determine the units digit for powers of any number efficiently?
2. Can you explain why the units digit of a number follows a repeating cycle?
3. What is the units digit of \(7^{2023}\)?
4. How would the result change if we summed powers of a different number, say 4 instead of 3?
5. Can you find the units digit of \(2013^{2013^{2013}}\)?

**Tip:** When working with large exponents, always look for repeating patterns in the units digits to simplify your calculations.

Discover how to find the units digit of the sum 2013^1 + 2013^2 + ... + 2013^2013 using modular arithmetic and cyclic patterns. This problem explores the units digits of powers of 2013 and demonstrates how to calculate the units digit efficiently. Suitable for high school students interested in number theory and modular arithmetic.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation