Final answer:
The given sum S is divisible by 15 by showing divisibility by 3 and 5 from the repeating pattern of powers of 2. S has a minimum of 30 digits as the largest term, 2⁹⁹, significantly exceeds a 10-digit number and thus the sum of all terms will as well.
Step-by-step explanation:
Proving divisibility by 15 and determining the number of digits in S
To address the question given, we have the following sum:
S = 2⁰ + 2¹ + 2² + 2³ + ........ + 2⁹⁹
(a) To prove that S is divisible by 15, notice that 15 is 3 times 5. We need to show that S is divisible by both these primes. For 3, the sum of powers of 2 ranging from 2⁰ to 2³ (1, 2, 4, 8) repeats every 4 terms and sums to 15, which is divisible by 3. For 5, since 2⁴ (16) ends in a 6 and every subsequent power of 2 will have a last digit that cycles through the same pattern (6, 2, 4, 8, 6...), adding these last digits will repeat every 4 terms, yielding a sum that is always divisible by 5. Therefore, the entire sum S is divisible by both 3 and 5, and thus by 15.
(b) To show that S has a minimum of 30 digits, we can look at the largest term in the sum which is 2⁹⁹. Since 2³⁰ is already a 10-digit number, 2⁹⁹, which is 2 raised to a power more than three times larger, will certainly have substantially more than 30 digits. Consequently, the sum S, which includes all powers of 2 up to 2⁹⁹, will also have a minimum of 30 digits.