Key Concepts Utilized

To solve this problem effectively, we will leverage the following mathematical concepts:

1. Sum of a Geometric Progression (GP): For a GP with first term $a$, common ratio $r$, and $n$ terms, the sum $S_n = \frac{a(r^n - 1)}{r-1}$ (when $r \neq 1$).
2. Modular Arithmetic: The system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value—the modulus. Key properties include:
   • If $a \equiv b \pmod m$ and $c \equiv d \pmod m$, then $a+c \equiv b+d \pmod m$ and $ac \equiv bd \pmod m$.
   • Solving linear congruences: $ax \equiv b \pmod m$ can be solved if $\gcd(a,m)$ divides $b$.
3. Euler's Totient Theorem: If $\gcd(a, n) = 1$, then $a^{\phi(n)} \equiv 1 \pmod n$, where $\phi(n)$ is Euler's totient function, which counts the number of positive integers up to $n$ that are relatively prime to $n$.
4. Chinese Remainder Theorem (CRT): Used to solve systems of linear congruences where the moduli are pairwise coprime. If $x \equiv a_1 \pmod{n_1}$, $x \equiv a_2 \pmod{n_2}$, ..., $x \equiv a_k \pmod{n_k}$, and $n_1, \dots, n_k$ are pairwise coprime, then there is a unique solution modulo $N = n_1 n_2 \dots n_k$.

Step 1: Calculate the Sum of the Geometric Progression

The given series is $1 + 3 + 3^2 + 3^3 + \dots + 3^{2021}$. This is a Geometric Progression with:

• First term ($a$) = 1
• Common ratio ($r$) = 3
• Number of terms ($n$) = $2021 - 0 + 1 = 2022$.

Using the formula for the sum of a GP, $S_n = \frac{a(r^n - 1)}{r-1}$: $S = \frac{1 \cdot (3^{2022} - 1)}{3 - 1}$ $S = \frac{3^{2022} - 1}{2}$ We need to find the remainder when $S$ is divided by 50, i.e., $S \pmod{50}$.

Step 2: Determine the Remainder using Modular Arithmetic and Chinese Remainder Theorem

To find $S \pmod{50}$, we can use the Chinese Remainder Theorem by finding the remainder of $S$ modulo the coprime factors of 50. Since $50 = 2 \times 25$, we will find $S \pmod 2$ and $S \pmod{25}$ separately.

Sub-step 2.1: Find $S \pmod 2$

The sum is $S = 1 + 3 + 3^2 + \dots + 3^{2021}$. Consider each term modulo 2. Since $3 \equiv 1 \pmod 2$, every term $3^k \equiv 1^k \equiv 1 \pmod 2$. $S \equiv (1 \pmod 2) + (1 \pmod 2) + (1 \pmod 2) + \dots + (1 \pmod 2)$ There are 2022 terms in the sum. $S \equiv 2022 \pmod 2$ Since 2022 is an even number, $2022 \equiv 0 \pmod 2$. Thus, $S \equiv 0 \pmod 2 \quad \text{(Equation 1)}$

Sub-step 2.2: Find $S \pmod{25}$

From Step 1, we have $S = \frac{3^{2022} - 1}{2}$. To find $S \pmod{25}$, we can write this as a congruence: $2S \equiv 3^{2022} - 1 \pmod{25}$ First, let's find $3^{2022} \pmod{25}$ using Euler's Totient Theorem. For $n=25 = 5^2$, Euler's totient function $\phi(25) = 25(1 - \frac{1}{5}) = 25 \times \frac{4}{5} = 20$. Since $\gcd(3, 25) = 1$, by Euler's Totient Theorem, $3^{20} \equiv 1 \pmod{25}$.

Now, we need to find the exponent $2022$ modulo $20$: $2022 = 20 \times 101 + 2$. So, $3^{2022} = 3^{20 \times 101 + 2} = (3^{20})^{101} \cdot 3^2$. $3^{2022} \equiv (1)^{101} \cdot 3^2 \pmod{25}$ $3^{2022} \equiv 1 \cdot 9 \pmod{25}$ $3^{2022} \equiv 9 \pmod{25}$ Substitute this back into the congruence for $2S$: $2S \equiv 9 - 1 \pmod{25}$ $2S \equiv 8 \pmod{25}$ To solve for $S$, we need to multiply by the modular inverse of $2 \pmod{25}$. The modular inverse of $2 \pmod{25}$ is $13$, because $2 \times 13 = 26 \equiv 1 \pmod{25}$. Multiply both sides by 13: $13 \times 2S \equiv 13 \times 8 \pmod{25}$ $26S \equiv 104 \pmod{25}$ Since $26 \equiv 1 \pmod{25}$ and $104 = 4 \times 25 + 4$, so $104 \equiv 4 \pmod{25}$: $S \equiv 4 \pmod{25} \quad \text{(Equation 2)}$

Sub-step 2.3: Combine results using Chinese Remainder Theorem

We have the system of congruences:

1. $S \equiv 0 \pmod 2$
2. $S \equiv 4 \pmod{25}$

From Equation 2, $S$ can be written in the form $S = 25k + 4$ for some integer $k$. Substitute this expression for $S$ into Equation 1: $25k + 4 \equiv 0 \pmod 2$ Consider the coefficients modulo 2: $25 \equiv 1 \pmod 2$ and $4 \equiv 0 \pmod 2$. $1k + 0 \equiv 0 \pmod 2$ $k \equiv 0 \pmod 2$ This means $k$ must be an even integer. Let $k = 2m$ for some integer $m$. Substitute $k = 2m$ back into the expression for $S$: $S = 25(2m) + 4$ $S = 50m + 4$ This implies that $S \equiv 4 \pmod{50}$.

Therefore, the remainder on dividing $1 + 3 + 3^2 + 3^3 + \dots + 3^{2021}$ by 50 is 4.

Important Tips / Common Mistakes

• Correct Number of Terms: Always double-check the number of terms in a series. For a series $a^0 + a^1 + \dots + a^n$, there are $n+1$ terms.
• Modular Inverse for Division: When solving congruences like $aX \equiv b \pmod m$, you cannot simply divide by $a$ unless $\gcd(a,m)=1$. If $\gcd(a,m)=1$, find the modular inverse of $a$ modulo $m$ and multiply both sides by it. If $\gcd(a,m) \neq 1$, you must use techniques like the Chinese Remainder Theorem or consider multiple solutions. In this case, $2S \equiv 8 \pmod{25}$ allowed division by finding the inverse of 2 mod 25, since $\gcd(2,25)=1$.
• Euler's Totient Theorem Condition: Remember that Euler's Totient Theorem $a^{\phi(n)} \equiv 1 \pmod n$ applies only when $\gcd(a, n) = 1$.
• Remainder Definition: The remainder when $A$ is divided by $M$ is an integer $R$ such that $0 \le R < M$.

Summary and Key Takeaway

This problem demonstrates a classic application of geometric series sum, modular arithmetic, Euler's Totient Theorem, and the Chinese Remainder Theorem. By breaking down the problem into finding remainders modulo coprime factors (2 and 25 for 50) and then combining the results, we can efficiently handle large exponents and divisions within modular arithmetic. The key is to systematically apply these theorems to simplify the calculations and arrive at the final remainder. The remainder on dividing $1 + 3 + 3^2 + \dots + 3^{2021}$ by 50 is $\boxed{4}$.