Finding Sequences of Five Consecutive Integers

The question asks us to find sequences of five consecutive integers such that their product is equal to one of the integers in the sequence. Let a sequence of five consecutive integers be represented by \(n, n+1, n+2, n+3, n+4\), where \(n\) is an integer. The product of these integers is \(P = n(n+1)(n+2)(n+3)(n+4)\).

According to the problem, the product \(P\) must be equal to one of the integers in the sequence. So, we need to solve the equation:

\[n(n+1)(n+2)(n+3)(n+4) = m\]

where \(m \in \{n, n+1, n+2, n+3, n+4\}\).

Analyzing the Product of Consecutive Integers

The product of five consecutive integers is zero if and only if one of the integers in the sequence is zero. Let's consider this case first.

• If \(n=0\), the sequence is \((0, 1, 2, 3, 4)\). The product is \(0 \times 1 \times 2 \times 3 \times 4 = 0\). Is \(0\) one of the integers in the sequence? Yes, it is the first integer (\(n\)). So, \((0, 1, 2, 3, 4)\) is a valid sequence.
• If \(n+1=0\), which means \(n=-1\), the sequence is \((-1, 0, 1, 2, 3)\). The product is \((-1) \times 0 \times 1 \times 2 \times 3 = 0\). Is \(0\) one of the integers in the sequence? Yes, it is the second integer (\(n+1\)). So, \((-1, 0, 1, 2, 3)\) is a valid sequence.
• If \(n+2=0\), which means \(n=-2\), the sequence is \((-2, -1, 0, 1, 2)\). The product is \((-2) \times (-1) \times 0 \times 1 \times 2 = 0\). Is \(0\) one of the integers in the sequence? Yes, it is the third integer (\(n+2\)). So, \((-2, -1, 0, 1, 2)\) is a valid sequence.
• If \(n+3=0\), which means \(n=-3\), the sequence is \((-3, -2, -1, 0, 1)\). The product is \((-3) \times (-2) \times (-1) \times 0 \times 1 = 0\). Is \(0\) one of the integers in the sequence? Yes, it is the fourth integer (\(n+3\)). So, \((-3, -2, -1, 0, 1)\) is a valid sequence.
• If \(n+4=0\), which means \(n=-4\), the sequence is \((-4, -3, -2, -1, 0)\). The product is \((-4) \times (-3) \times (-2) \times (-1) \times 0 = 0\). Is \(0\) one of the integers in the sequence? Yes, it is the fifth integer (\(n+4\)). So, \((-4, -3, -2, -1, 0)\) is a valid sequence.

So, any sequence of five consecutive integers that includes 0 is a valid sequence.

Considering Cases Where the Product is Non-Zero

For the product to be non-zero, none of the integers in the sequence can be zero. This means either \(n > 0\) or \(n+4 < 0\) (i.e., \(n < -4\)).

• Case 1: \(n > 0\) (\(n \ge 1\)) The integers are \(n, n+1, n+2, n+3, n+4\). All are positive integers \(\ge 1\). The product \(P = n(n+1)(n+2)(n+3)(n+4)\). The integers in the sequence are \(n, n+1, n+2, n+3, n+4\). The largest is \(n+4\). For \(n=1\), the sequence is \((1, 2, 3, 4, 5)\). The product is \(1 \times 2 \times 3 \times 4 \times 5 = 120\). The integers are \(\{1, 2, 3, 4, 5\}\). Is \(120\) equal to any of these? No. For \(n \ge 1\), the product \(n(n+1)(n+2)(n+3)(n+4)\) grows much faster than \(n+4\). For \(n=1\), \(120 > 5\). For \(n \ge 1\), \(n \ge 1\), \(n+1 \ge 2\), \(n+2 \ge 3\), \(n+3 \ge 4\), \(n+4 \ge 5\). The product \(P \ge 1 \times 2 \times 3 \times 4 \times 5 = 120\). The largest integer in the sequence is \(n+4\). For \(n=1\), \(n+4=5\). \(120 > 5\). For \(n=2\), \(n+4=6\). \(2 \times 3 \times 4 \times 5 \times 6 = 720 > 6\). For any \(n \ge 1\), \(n(n+1)(n+2)(n+3)(n+4) > n+4\). To see this, divide the product by \(n+4\) (assuming \(n+4 \neq 0\), which is true for \(n \ge 1\)): \(n(n+1)(n+2)(n+3)\). This is a product of four positive consecutive integers, which is always greater than 1 for \(n \ge 1\). For example, if \(n=1\), \(1(2)(3)(4) = 24 > 1\). If \(n=1\), \(n(n+1)(n+2)(n+3)(n+4) = 120\), and \(n+4=5\). \(120 \neq 5\). If the product \(P\) equals \(n, n+1, n+2\), or \(n+3\), it would also be greater than these values for \(n \ge 1\). Thus, there are no valid sequences when \(n > 0\).
• Case 2: \(n < -4\) (\(n \le -5\)) The integers are \(n, n+1, n+2, n+3, n+4\). All are negative integers \(\le -1\). The product \(P = n(n+1)(n+2)(n+3)(n+4)\). Since there are five negative terms, the product is negative. The integers in the sequence are \(\{n, n+1, n+2, n+3, n+4\}\). The largest is \(n+4\) (least negative), and the smallest is \(n\) (most negative). For \(n=-5\), the sequence is \((-5, -4, -3, -2, -1)\). The product is \((-5)(-4)(-3)(-2)(-1) = -120\). The integers are \(\{-5, -4, -3, -2, -1\}\). Is \(-120\) equal to any of these? No. The smallest integer in the sequence is -5. The product -120 has a much larger magnitude than -5. For \(n \le -5\), let's consider the magnitude of the product: \(|n(n+1)(n+2)(n+3)(n+4)| = |n||n+1||n+2||n+3||n+4|\). Since \(n \le -5\), \(|n| \ge 5\), \(|n+1| \ge 4\), \(|n+2| \ge 3\), \(|n+3| \ge 2\), \(|n+4| \ge 1\). The magnitude of the product \(|P| \ge 5 \times 4 \times 3 \times 2 \times 1 = 120\). The magnitudes of the integers in the sequence are \(|n|, |n+1|, |n+2|, |n+3|, |n+4|\). Since \(n \le -5\), these are \(|n|\), \(|n|-1\), \(|n|-2\), \(|n|-3\), \(|n|-4\). The largest magnitude is \(|n|\). We need \(P = m\) where \(m \in \{n, n+1, n+2, n+3, n+4\}\). Since \(P\) is negative and the integers in the sequence are negative, we need \(P = m\) for some \(m\). However, for \(n \le -5\), the magnitude of the product is \(|P| \ge 120\). The magnitude of any integer in the sequence is \(|n+i|\) for \(i \in \{0, 1, 2, 3, 4\}\). The largest magnitude is \(|n|\) when \(i=0\). For \(n=-5\), \(|n|=5\). For \(n=-6\), \(|n|=6\). For \(n \le -5\), \(|n(n+1)(n+2)(n+3)(n+4)| \ge 120\). The maximum magnitude of an integer in the sequence is \(|n|\). For \(|n| \ge 5\), \(120 > |n|\). Therefore, the magnitude of the product is always greater than the magnitude of any integer in the sequence for \(n \le -5\). Since both the product and the integers in the sequence are negative, the product cannot equal any of the integers. Thus, there are no valid sequences when \(n < -4\).

Listing the Valid Sequences

Based on the analysis, the only valid sequences are those containing 0. These sequences (collection X) are:

1. \((0, 1, 2, 3, 4)\)
2. \((-1, 0, 1, 2, 3)\)
3. \((-2, -1, 0, 1, 2)\)
4. \((-3, -2, -1, 0, 1)\)
5. \((-4, -3, -2, -1, 0)\)

Identifying P and Q

P is the smallest integer occurring in these sequences. Let's list all integers present in these sequences:

\(\{0, 1, 2, 3, 4\} \cup \{-1, 0, 1, 2, 3\} \cup \{-2, -1, 0, 1, 2\} \cup \{-3, -2, -1, 0, 1\} \cup \{-4, -3, -2, -1, 0\}\)

The set of all distinct integers appearing in these sequences is \(\{-4, -3, -2, -1, 0, 1, 2, 3, 4\}\).

The smallest integer in this set is \(P = -4\).

The largest integer in this set is \(Q = 4\).

Calculating the Arithmetic Mean of P and Q

The arithmetic mean of P and Q is given by the formula:

\[\text{Arithmetic Mean} = \frac{P+Q}{2}\]

Substituting the values \(P=-4\) and \(Q=4\):

\[\text{Arithmetic Mean} = \frac{-4 + 4}{2} = \frac{0}{2} = 0\]

The arithmetic mean of P and Q is 0.

The collection X of all possible such sequences is \(\{ (0, 1, 2, 3, 4), (-1, 0, 1, 2, 3), (-2, -1, 0, 1, 2), (-3, -2, -1, 0, 1), (-4, -3, -2, -1, 0) \}\).

The integers occurring in these sequences are \(\{-4, -3, -2, -1, 0, 1, 2, 3, 4\}\).

The smallest integer P is -4.

The largest integer Q is 4.

Arithmetic mean of P and Q \(= \frac{-4 + 4}{2} = \frac{0}{2} = 0\).

Revision Table: Key Concepts

Additional Information: Properties of Consecutive Integer Products

The product of \(k\) consecutive integers is always divisible by \(k!\). For five consecutive integers, the product \(n(n+1)(n+2)(n+3)(n+4)\) is always divisible by \(5! = 120\).

We found that for sequences not containing 0, the magnitude of the product grows very rapidly. For \(n \ge 1\), the product is \(\ge 120\), while the largest element is \(n+4\). For \(n \le -5\), the product is \(\le -120\), while the smallest element is \(n\). The magnitude of the product quickly exceeds the magnitude of any element in the sequence, preventing the product from being equal to an element.

The special case where the product is 0 (because 0 is in the sequence) leads to the product equalling one of the elements (0 itself), which is why these sequences are the only solutions.