Question

Set T is a finite set of positive consecutive multiples of 14. How many of these integers are also multiples of 21?
1. Set T consists of 30 integers.
2. The smallest integer in Set T is a multiple of 21.

• A. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked
• B. EACH statement ALONE is sufficient to answer the question asked
• C. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed
• D. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked
• E. Both statements (1) and (2) TOGETHER are sufficient to answer the question asked; but NEITHER statement ALONE is sufficient

Hint:

In data sufficiency questions involving multiples, always find the LCM of the given numbers. For consecutive items, the number of multiples of 'n' in a set of 'k' consecutive items is either \(\lfloor k/n \rfloor\) or \(\lceil k/n \rceil\). If k is a multiple of n, the count is exactly \(k/n\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The question asks for the number of integers in a set T that are multiples of both 14 and 21. An integer that is a multiple of both 14 and 21 must be a multiple of their least common multiple (LCM).

Step 2: Key Formula or Approach:
First, we find the LCM of 14 and 21.
The prime factorization of 14 is \(2 \times 7\).
The prime factorization of 21 is \(3 \times 7\).
The LCM is the product of the highest powers of all prime factors that appear in either factorization.
\[ \text{LCM}(14, 21) = 2^1 \times 3^1 \times 7^1 = 42 \] So, the question is equivalent to asking: "How many multiples of 42 are in Set T?"
Set T consists of positive consecutive multiples of 14. We can represent the elements of T as \(14k, 14(k+1), 14(k+2), \dots\) for some positive integer \(k\).
For an element \(14m\) to be a multiple of 42, we must have \(14m = 42j\) for some integer \(j\).
Dividing by 14, we get \(m = 3j\). This means that a member of set T is a multiple of 42 if and only if its corresponding multiplier is a multiple of 3.

Step 3: Detailed Explanation:
Analyzing Statement (1): Set T consists of 30 integers.
The set T can be written as \(\{14k, 14(k+1), \dots, 14(k+29)\}\) for some positive integer \(k\).
We need to find how many of these elements are multiples of 42. This is equivalent to finding how many of the consecutive integers \(\{k, k+1, \dots, k+29\}\) are multiples of 3.
In any set of 30 consecutive integers, there will always be exactly \(30 \div 3 = 10\) multiples of 3.
For example, if \(k=1\), the multipliers are \(\{1, 2, ..., 30\}\), which contains multiples of 3: 3, 6, 9, ..., 30 (10 numbers).
If \(k=2\), the multipliers are \(\{2, 3, ..., 31\}\), which contains multiples of 3: 3, 6, 9, ..., 30 (10 numbers).
If \(k=3\), the multipliers are \(\{3, 4, ..., 32\}\), which contains multiples of 3: 3, 6, 9, ..., 30 (10 numbers).
Thus, there are always 10 integers in the set T that are multiples of 42.
Statement (1) alone is sufficient to answer the question.
Analyzing Statement (2): The smallest integer in Set T is a multiple of 21.
Let the smallest integer be \(14k\). We are given that \(14k\) is a multiple of 21.
This means \(14k\) must be a multiple of LCM(14, 21), which is 42.
So, the first element of the set is a multiple of 42. The set looks like \(\{42j, 42j+14, 42j+28, 42j+42, \dots \}\).
The multiples of 42 appear every 3 terms (the 1st term, 4th term, 7th term, etc.).
However, we do not know the size of the set T. If the set has 3 elements, there is 1 multiple of 42. If the set has 5 elements, there is still 1 multiple of 42. If the set has 6 elements, there are 2 multiples of 42.
Since we cannot determine a unique number of multiples of 42, statement (2) alone is not sufficient.

Step 4: Final Answer:
Statement (1) alone is sufficient, but statement (2) alone is not sufficient.