Final answer:
A to the power of 21, which represents a number with 21 consecutive ones, is not divisible by 11 since it has an odd number of ones but is divisible by 13 because the pattern shows that a number with repetitive ones that is a multiple of 3 in length, like 111111, is divisible by 13.
Step-by-step explanation:
The student's question revolves around the divisibility of a large number composed of repeated 1's, specifically A to the power of 21, which represents the number 111...111 (with 21 ones). To check the divisibility of this number by the choices given, it's necessary to understand repetition of digits and divisibility rules.
It is known that a number comprising of repetitive 1's in its digits, like 111 or 11111, has unique divisibility properties. A quick way to investigate is to actually construct the smallest such numbers for each divider in options A, B, C, and D and see the pattern:
- 11, 111, 1111, etc. (divisible by 11)
- 111, 111111, etc. (divisible by 3, 37, and therefore by 111)
Upon examining the pattern, it becomes apparent that for 11, a number formed by repeating units is divisible by 11 if it has an even number of ones. Since the power of 21 means 21 ones, which is odd, A to the power of 21 will not be divisible by 11. However, if we look at the pattern for divisibility by 13, we find that 111111 is divisible by 13, and since 21 is a multiple of 3, it means that A to the power of 21 would equally be divisible. Hence, the correct answer from the given options would be B) 13.