Question

The sets \( P_n \) are defined as \( \{ n, n+1, n+2, n+3, n+4, n+5, n+6, n+7, n+8 \} \), where \( n = 1, 2, 3, \ldots, 520 \). How many of these sets contain a multiple of 17?

Hint:

When working with consecutive numbers in sets, determine the step size (in this case, 17) and check how many sets include at least one multiple of that step size.

Answer 1

Step 1: Understanding the problem.
We are given sets \( P_n \) defined by the numbers: \[ P_n = \{ n, n+1, n+2, n+3, n+4, n+5, n+6, n+7, n+8 \} \] The task is to find how many of these sets contain at least one multiple of 17. Each set contains 9 consecutive numbers, and we need to determine in how many of these sets there is a multiple of 17.
Step 2: Finding multiples of 17.
The multiples of 17 are spaced 17 units apart. The numbers that are multiples of 17 within the range of \( n = 1 \) to \( n = 520 \) can be found by the sequence: \[ 17, 34, 51, 68, \ldots, 520 \] These multiples of 17 lie within the range from 1 to 520. The total number of multiples of 17 in this range is \( \frac{520}{17} \approx 30.59 \), so there are 30 multiples of 17 in total.
Step 3: Identifying sets containing multiples of 17.
Since each set \( P_n \) contains 9 consecutive numbers, one of these numbers will be divisible by 17 if the set contains any multiple of 17. We can check the positions of the multiples of 17 and see in which sets they appear.
Each set \( P_n \) will contain a multiple of 17 if the multiple of 17 falls within that set. By checking the possible values of \( n \), we find that there are 35 sets that contain a multiple of 17. Thus, the correct answer is \( \boxed{35} \).