Question

If \(n = 1 + x\), where \(x\) is the product of 4 consecutive positive integers, then which of the following is/are true?
1. \(n\) is odd
2. \(n\) is prime
3. \(n\) is a perfect square

• A. 1 and 3 only
• B. 1 and 2 only
• C. 1 only
• D. None of these

Hint:

Write \(x\) as \((k^2+3k)(k^2+3k+2)\) and notice this has the form \(m(m+2)\), so \(x+1\) is a perfect square.

Answer 1

The Correct Option is A

A quick way into this question is to test a few small cases of "4 consecutive positive integers" and look for a pattern, then confirm that pattern with a short piece of algebra so it holds for every case, not just the ones tested.

Take $k=1,2,3,4$ and compute $x$ and $n=x+1$ each time:

• $k=1$: $x = 1 \times 2 \times 3 \times 4 = 24$, so $n = 25 = 5^2$.
• $k=2$: $x = 2 \times 3 \times 4 \times 5 = 120$, so $n = 121 = 11^2$.
• $k=3$: $x = 3 \times 4 \times 5 \times 6 = 360$, so $n = 361 = 19^2$.
• $k=4$: $x = 4 \times 5 \times 6 \times 7 = 840$, so $n = 841 = 29^2$.

Every value of $n$ came out odd and also a perfect square, and none of them were prime (25, 121, 361, and 841 are all composite). This strongly suggests statements 1 and 3 are always true and statement 2 is always false, but a pattern from four examples is not a proof, so it helps to see why algebraically.

1. Statement 1, $n$ is odd: any run of 4 consecutive integers contains at least two even numbers, so their product $x$ is always even. Adding 1 to an even number gives an odd number, so $n=x+1$ is always odd.
2. Statement 2, $n$ is prime: writing $x = k(k+1)(k+2)(k+3) = (k^2+3k)(k^2+3k+2)$ and setting $m=k^2+3k$, we get $x = m^2+2m$, so $n = x+1 = (m+1)^2$. Since $n$ is a perfect square greater than 1, it always has the factor pair $1$ and $n$ plus the middle factor $m+1$, giving it at least three divisors. A number with more than two divisors is never prime, so statement 2 always fails.
3. Statement 3, $n$ is a perfect square: the same identity $n=(m+1)^2$ shows this directly. Whatever whole number $m+1$ turns out to be, $n$ is its square, so statement 3 always holds.

So statements 1 and 3 are always true for every choice of 4 consecutive positive integers, and statement 2 is never true, since a perfect square bigger than 1 cannot be prime.

Let's summarize:

• $x$, the product of 4 consecutive integers, is always even, making $n=x+1$ always odd.
• $x+1$ always simplifies to $(m+1)^2$ where $m=k^2+3k$, so $n$ is always a perfect square.
• A perfect square greater than 1 always has more than two divisors, so $n$ is never prime.

The correct choice is "1 and 3 only", option (A).