57. What is the remainder when 9³ + 9⁴ + 9⁵ + 9⁶ + … + 9¹⁰⁰ is divided by 6?

1. B. 1
2. C. 2
3. D. 3

Answer: A

Explanation

We need to find the remainder when the sum S = 9³ + 9⁴ + 9⁵ + … + 9¹⁰⁰ is divided by 6.
Let’s look at the remainder of 9 raised to any power when divided by 6:
* 9 mod 6 = 3
* 9² mod 6 = (9 × 9) mod 6 = (3 × 3) mod 6 = 9 mod 6 = 3
* 9³ mod 6 = (9² × 9) mod 6 = (3 × 3) mod 6 = 9 mod 6 = 3
In general, for any integer k ≥ 1, 9ᵏ mod 6 = 3.

Now, the sum S = 9³ + 9⁴ + 9⁵ + … + 9¹⁰⁰.
There are (100 – 3) + 1 = 98 terms in the sum.
Each term 9ᵏ (for k from 3 to 100) has a remainder of 3 when divided by 6.
So, S mod 6 = (3 + 3 + 3 + … + 3) mod 6 (98 times)
S mod 6 = (98 × 3) mod 6
98 × 3 = 294.
294 mod 6 = 0 (since 294 = 49 × 6).

Therefore, the remainder is 0. This question tests number theory concepts, specifically remainders and modular arithmetic, which are common in CSAT.