Introduction

Understanding the last digit of large powers, such as 72016, can help us apply various mathematical theorems and patterns. This article will explore different methods to find the last digit of this number, including the cyclicity of the unit digit of 7 and Euler's Theorem.

Selecting a Pattern Approach

To find the last digit of 72016, we begin by observing a pattern in the last digits of the first few powers of 7:

71 7, last digit is 7 72 49, last digit is 9 73 343, last digit is 3 74 2401, last digit is 1

Notice that the last digits repeat every four powers: 7, 9, 3, 1. To determine the last digit of 72016, we calculate 2016 mod 4:

2016 div 4  504 remainder 0

Since the remainder is 0, 72016 corresponds to the last digit of 74, which is 1. Therefore, the last digit of 72016 is 1.

Using Euler's Theorem

Euler's Theorem states that if (a) and (n) are coprime, then aφ(n) ≡ 1 (mod n). Here, 7 and 10 are coprime, and φ(10) 4. Therefore, (7^4 ≡ 1 pmod{10}).

Given that 2016 4 × 504, we can express 72016 as (74)504. Using Euler's Theorem, we get:

72016  (74)504 ≡ 1504 ≡ 1 (mod 10)

Hence, the last digit is 1.

Another Pattern Approach

We can also observe the last digits directly:

Last digit of 71 is 7 Last digit of 72 is 9 Last digit of 73 is 3 Last digit of 74 is 1

Since 2016 is a multiple of 4, the last digit of 72016 is the same as the last digit of 74, which is 1.

Using Euler’s Theorem in a More Complex Way

Let's consider a more complex use of Euler's Theorem for 72017 and then apply it to 72016 for completeness:

(varphi(10) 4), because (10 2 cdot 5) and (varphi(2 cdot 5) (2-1)(5-1) 4) Hence, (7^4 ≡ 1 pmod{10}). Now, (2017 4 cdot 504 1), so 72017 has a last digit of 7. For 72016, we use 2016 4 cdot 504, so (7^{2016} (7^4)^{504} ≡ 1^{504} ≡ 1 pmod{10}).

The last digit is 1.

Conclusion

To summarize, the last digit of 72016 can be found using either the pattern of last digits or Euler's Theorem. Both methods lead to the same result: the last digit is 1.