What is the remainder when the sum S = 3^1 + 3^2 + 3^3 + \dots + 3^{2025} is divided by 10… — GMAT

What is the remainder when the sum $S = 3^1 + 3^2 + 3^3 + \dots + 3^{2025}$ is divided by $100$ ?

Answer: 63

The remainder when the sum

S

is divided by

100

63

Using the geometric series formula,

S = \frac{3(3^{2025} - 1)}{2}

. Computing

3^{2025} \pmod{200}

gives

43 \pmod{200}

because

3^{20} \equiv 1 \pmod{200}

and

2025 \equiv 5 \pmod{20}

. Substituting this into the numerator yields

3(43 - 1) = 126 \pmod{200}

. Dividing by

2

produces the remainder

63

modulo

100

Sum the geometric series

S = 3^1 + 3^2 + \dots + 3^{2025}

S = \frac{3(3^{2025} - 1)}{2}

Applying the standard finite geometric series formula

a\frac{r^n - 1}{r - 1}

with

a = 3

and

r = 3

Set up modular equation for

S \pmod{100}

Evaluate

3(3^{2025} - 1) \pmod{200}

Dividing an integer congruence by

2

requires doubling the modulus to

200

to maintain exact divisibility.

Find the cyclicity period of

3^n \pmod{200}

3^{20} \equiv 1 \pmod{200}

3^5 = 243 \equiv 43

3^{10} \equiv 43^2 = 1849 \equiv 49

, and

3^{20} \equiv 49^2 = 2401 \equiv 1 \pmod{200}

Compute

3^{2025} \pmod{200}

3^{2025} \equiv 3^5 \equiv 43 \pmod{200}

Dividing the exponent

2025

by the period length

20

leaves a remainder of

5

Calculate the numerator

3(3^{2025} - 1) \pmod{200}

and divide by 2

3(43 - 1) = 126 \pmod{200}

, so

S \equiv 63 \pmod{100}

Simplifying

126 / 2 = 63

gives the unique remainder modulo

100

Geometric series summation combined with modular arithmetic and cyclicity mod 200

Question