Question

Consider the set S = 1, 2, 3, …, 10001. How many arithmetic progressions with at least 3 elements can be formed from the elements of S that start with 1 and end with 1000?

• A. 6
• B. 7
• C. 8
• D. 12
• E. 13

Answer 1

The Correct Option is B

Let's find the number of arithmetic progressions (APs) that can be formed from the set \( S = \{1, 2, 3, \ldots, 10001\} \) with at least 3 elements that start with 1 and end with 1000. Here are the steps to determine this: 

1. First, understand that an arithmetic progression is a sequence of numbers with a common difference, \( d \). So, the terms can be represented as \( a, a+d, a+2d, \ldots \).
2. Here, the first term is fixed at 1, and the last term in the sequence must be 1000. Therefore, any AP must satisfy:
   \(1 + nd = 1000\) where \( n \) is the number of elements in the AP minus one.
3. Solve for \( d \):
   \(nd = 999\)
   We need \( n \) and \( d \) such that the set has at least 3 elements, meaning \( n \geq 2 \), and \( d \) is a divisor of 999.
4. Determine the divisors of 999:
   • The number 999 can be factored as \( 999 = 3^3 \times 37 \).
   • The divisors of 999 are: 1, 3, 9, 27, 37, 111, 333, 999.
   • Since \( d \) must be such that there are at least 3 elements, consider values where \( n \geq 2 \).
5. Check potential values for \( d \):
   • For each divisor \( d \), compute \( n = \frac{999}{d} \).
   • The number of terms in the AP is \( n + 1 \), which must be ≥ 3, implying \( n \) must be ≥ 2.
   • Check all divisors:
     • If \( d = 999 \), \( n = 1 \) – invalid (only 2 terms).
     • If \( d = 333 \), \( n = 3 \) – valid (3 terms: 1, 334, 667, 1000).
     • If \( d = 111 \), \( n = 9 \) – valid (more terms).
     • If \( d = 37 \), \( n = 27 \) – valid.
     • If \( d = 27 \), \( n = 37 \) – valid.
     • If \( d = 9 \), \( n = 111 \) – valid.
     • If \( d = 3 \), \( n = 333 \) – valid.
     • If \( d = 1 \), \( n = 999 \) – valid.
6. Each of these valid cases meets the requirement of \( n \geq 2 \). Thus, there are 7 valid divisors: 1, 3, 9, 27, 37, 111, and 333.

Therefore, there are 7 arithmetic progressions that can be formed from the elements of set \( S \) starting with 1 and ending with 1000.