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

To find the remainder when 9³ + 9⁴ + 9⁵ + 9⁶ + … + 9¹⁰⁰ is divided by 6, let's analyze the remainder of each term when divided by 6. 1. Remainder of 9 when divided by 6: 9 = 1 * 6 + 3. So, 9 leaves a remainder of 3 when divided by 6. We can write this as 9 ≡ 3 (mod 6). 2. Remainder of powers of 9 when divided by 6: If 9 ≡ 3 (mod 6), then 9ⁿ ≡ 3ⁿ (mod 6). Let's check the powers of 3 modulo 6: 3¹ ≡ 3 (mod 6) 3² = 9 ≡ 3 (mod 6) 3³ = 27 ≡ 3 (mod 6) In general, for any integer n ≥ 1, 3ⁿ ≡ 3 (mod 6). Therefore, each term 9ⁿ in the sum (where n ≥ 3) will leave a remainder of 3 when divided by 6. 3. Number of terms in the sum: The exponents range from 3 to 100. Number of terms = (Last exponent - First exponent) + 1 Number of terms = (100 - 3) + 1 = 97 + 1 = 98 terms. 4. Sum of the remainders: The sum S = 9³ + 9⁴ + 9⁵ + 9⁶ + … + 9¹⁰⁰. When S is divided by 6, the remainder will be the sum of the individual remainders, modulo 6. S ≡ (3 + 3 + 3 + ... + 3) (mod 6) Since there are 98 terms, this is: S ≡ (98 * 3) (mod 6) 5. Calculate (98 * 3) mod 6: 98 * 3 = 294. Now, divide 294 by 6: 294 ÷ 6 = 49. Since 294 is perfectly divisible by 6, the remainder is 0. Alternatively, to find (98 * 3) mod 6: Notice that 98 is an even number, so 98 * 3 is a multiple of 2. Also, 98 * 3 is a multiple of 3. Since 98 * 3 is a multiple of both 2 and 3, it must be a multiple of their least common multiple, which is 6. Therefore, (98 * 3) mod 6 = 0. The remainder when 9³ + 9⁴ + 9⁵ + 9⁶ + … + 9¹⁰⁰ is divided by 6 is 0. The final answer is A