Final answer:
To find the number of positive integers n, no greater than 100, for which the sum of the first n odd numbers is divisible by 9, we can use sum formulas. The sum of the first n odd numbers can be represented as n^2. Therefore, there are 3 values of n (2, 6, and 9) for which the sum of the first n odd numbers is divisible by 9.
Step-by-step explanation:
To find the number of positive integers n, no greater than 100, for which the sum of the first n odd numbers is divisible by 9, we can use sum formulas.
The sum of the first n odd numbers can be represented as n^2. So we need to find the values of n for which n^2 is divisible by 9.
We know that 9 is a perfect square and its factors are 1, 3, and 9. Therefore, n^2 is divisible by 9 when n is a multiple of 3.
Dividing 100 by 3 gives us 33 with a remainder of 1. So, there are 33 numbers that are multiples of 3 between 1 and 100. However, we need to count the numbers that are divisible by 3 more than once (e.g., 9, 15, 21, etc.).
These numbers are the perfect squares of the multiples of 3, so we need to find the perfect squares that are less than or equal to 100. The perfect squares that are less than or equal to 100 and multiples of 3 are 9, 36, and 81.
Therefore, there are 3 values of n (2, 6, and 9) for which the sum of the first n odd numbers is divisible by 9. Hence, the answer is A. 3.