Introduction to Finding the Last Two Digits of (2^{5^k}) and Its Sum

In this article, we explore the mathematical techniques needed to find the last two digits of the sequence (2^{5^k}) and its sum. This problem is a delightful application of modular arithmetic and Euler's theorem. We will break down the steps and prove the patterns using mathematical induction and properties of exponents.

Understanding the Problem and Key Concepts

The main objective is to determine the last two digits of the sum of the sequence (S sum_{k1}^{2015} 2^{5^k}). This requires us to find the value of (2^{5^k} mod 100).

Pattern Identification Using Modular Arithmetic

In the field of modular arithmetic, patterns in the last digits of powers of numbers can often be found and utilized. We start by examining the last two digits of powers of 2, specifically (2^n mod 100), which repeat in a cycle of 20 starting at (2^2).

04 08 16 32 64 28 56 12 24 4896 92 84 68 36 72 44 88 76 52 04 08 ...

From this cycle, we observe that (2^{5} equiv 32 mod 100). This implies that the last two digits of (2^{5^n}) are 32 for all positive integers (n).

Proof by Induction

To formally establish that (2^{5^n} equiv 32 mod 100) for all positive integers (n), we use mathematical induction.

Base Case: For (n 1), (2^{5^1} 2^5 32 equiv 32 mod 100). Inductive Step: Assume the statement is true for (n m), i.e., (2^{5^m} equiv 32 mod 100). We need to prove it for (n m 1).

Starting with (2^{5^{m 1}} 2^{5 cdot 5^m}), we use the fact that (2^{20} equiv 1 mod 25) and the relationship between 20 and 100, specifically that (2^{5^m} equiv 32 mod 100).

[2^{5^{m 1}}  2^{5 cdot 5^m} equiv (32)^5 equiv (30   2)^5 mod 100]

Using the binomial theorem, we simplify ((30 2)^5), and find that the expression reduces to (equiv 32 mod 100).

Hence, by induction, (2^{5^n} equiv 32 mod 100) for all positive integers (n).

Calculating the Sum

With the established pattern, the problem simplifies to finding the last two digits of the sum of 2015 terms, each equal to 32. This can be represented as:

[sum_{k1}^{2015} 2^{5^k} equiv 2015 cdot 32 mod 100]

Simplifying the sum further:

[2015 cdot 32 equiv 15 cdot 32 equiv 480 mod 100 equiv 80 mod 100]

Thus, the last two digits of the sum are 80.

Conclusion

In summary, we have utilized modular arithmetic and induction to solve the problem of finding the last two digits of the sequence (2^{5^k}) and its sum. The result, as confirmed by our step-by-step analysis and proof, is that the last two digits of the sum are 80.