Products of consecutive integers

Pick any two whole numbers standing next to each other on the number line — say $17$ and $18$ — and multiply them. No matter which pair you choose, the answer is always even. Pick three neighbours in a row, like $8, 9, 10$ , and multiply all three, and the answer is always divisible by $6$ . These are not coincidences of the examples chosen; they are guaranteed properties that follow from how evenness and multiples of $3$ are spaced out along the integers. This topic lives inside the Number Play chapter of Class 8 Mathematics, and it is one of the cleanest places to see parity and divisibility reasoning turned into a genuine proof. The tools are simple — writing consecutive integers as $n, n+1, n+2, \ldots$ and arguing by cases — but the payoff is a statement that holds for every choice of starting number, not just the ones you happened to test.

The core insight

Among any $k$ consecutive integers, one is always a multiple of $2$ (for $k \geq 2$ ) and one is always a multiple of $3$ (for $k \geq 3$ ), which is why the product of two consecutive integers is always even and the product of three consecutive integers is always divisible by $6$ .

The core idea

Consecutive integers are numbers that follow one another with no gaps, written algebraically as $n, n+1, n+2, \ldots$ for whatever whole number $n$ you start at. The whole argument in this topic rests on one spacing fact: multiples of $2$ occur every second number, and multiples of $3$ occur every third number.

Two consecutive integers, $n$ and $n+1$ . Because they differ by exactly $1$ , they can never both be even or both be odd — one is always even. Since even $\times$ anything is even, the product is guaranteed to be even:

n(n+1) \equiv 0 \pmod{2} \text{ for every whole number } n.

where: $n$ is any whole number, and the product $n(n+1)$ denotes two consecutive integers multiplied together.

Three consecutive integers, $n$ , $n+1$ , $n+2$ . Any two neighbours inside this trio already supply a factor of $2$ , by the same reasoning as above. Separately, because multiples of $3$ occur every third whole number ( $3, 6, 9, \ldots$ ), any block of three consecutive integers must contain exactly one multiple of $3$ — it cannot skip past three numbers in a row without landing on one. Since $2$ and $3$ share no common factor ( $\gcd(2,3)=1$ ), having both factors present means the product is divisible by their combination:

n(n+1)(n+2) \equiv 0 \pmod{6} \text{ for every whole number } n.

where: $6 = 2 \times 3$ , and the congruence means the product leaves remainder $0$ when divided by $6$ . The reasoning generalises informally: the product of $k$ consecutive integers is always divisible by $k!$ (so four consecutive integers are always divisible by $24$ ), but Class 8 only needs the $k=2$ and $k=3$ cases, justified by cases rather than by formal modular arithmetic.

Figure

A number line marking three consecutive integers eight nine and ten with the multiple of two boxes and the multiple of three circled

Derivation AWhy three consecutive integers are always divisible by six

The goal is to show, for any three consecutive whole numbers $n, n+1, n+2$ with no exceptions and no need to check individual cases by trial, that their product $n(n+1)(n+2)$ is always divisible by $6$ , by separately establishing divisibility by $2$ and by $3$ and then combining the two facts.

Step 1. Establish divisibility by $2$ . Among $n$ and $n+1$ (a sub-pair already inside the trio), one is always even, since consecutive integers alternate between even and odd. An even number multiplied by anything stays even, so $n(n+1)(n+2)$ is divisible by $2$ .

Step 2. Establish divisibility by $3$ . Every whole number leaves remainder $0$ , $1$ , or $2$ when divided by $3$ . Whichever remainder $n$ leaves, checking $n$ , $n+1$ , $n+2$ in turn shows that exactly one of the three lands on a remainder of $0$ — because moving from $n$ to $n+1$ to $n+2$ steps through all three possible remainders in order, wrapping around at most once. So one of the three factors is always a multiple of $3$ .

Step 3. Combine the two divisibility facts. The product $n(n+1)(n+2)$ is divisible by $2$ (Step 1) and by $3$ (Step 2). Since $2$ and $3$ have no common factor other than $1$ (that is, $\gcd(2,3)=1$ ), a number divisible by both must be divisible by their product, $2 \times 3 = 6$ .

Step 4. Confirm the argument is general. Nothing in Steps 1–3 depended on the specific value of $n$ — the argument holds for $n=1$ just as it holds for $n=100$ or $n=8$ . This is why the claim is a proof, not a pattern noticed from a few examples.

Worked example

Prove that $8 \times 9 \times 10$ is divisible by $6$ , and verify the result by direct division.

Step 1. Identify the three consecutive integers: $n=8$ , $n+1=9$ , $n+2=10$ .

Step 2. Check divisibility by $2$ . Among $8$ and $10$ , both happen to be even, so the trio is certainly divisible by $2$ (any one even factor is already enough).

Step 3. Check divisibility by $3$ . Among $8, 9, 10$ , exactly one — namely $9 = 3 \times 3$ — is a multiple of $3$ , confirming the "one multiple of $3$ per block of three" claim.

Step 4. Combine. Since the product is divisible by both $2$ and $3$ , it is divisible by $6$ .

Step 5. Verify directly. $8 \times 9 \times 10 = 720$ , and $720 \div 6 = 120$ exactly, with no remainder.

The product $8\times9\times10 = 720$ is divisible by 6, as guaranteed.

Competitive stretch

$720$ is in fact divisible by far more than $6$ — $720 = 6 \times 120$ — because $8 = 2^3$ contributes extra factors of $2$ beyond the guaranteed minimum. This does not break the rule: "always divisible by $6$ " promises a minimum, and specific number choices are free to carry additional factors on top of it.

Common traps

Treating "always divisible by $6$ " as "the product is exactly $6$ times something small" — the guarantee is a minimum divisibility, and the actual quotient can be much larger, as with $720 = 6 \times 120$ .
Confusing the spacing rule for multiples of $2$ (every second number) with the spacing rule for multiples of $3$ (every third number) — the two arguments must be made separately, not merged into one.
Checking only one or two numerical examples and calling that a proof — a claim about all consecutive integers needs the general algebraic argument with $n, n+1, n+2$ , not a handful of successful trials.
Forgetting to identify which factor supplies the $2$ and which supplies the $3$ when asked to justify an answer — stating the conclusion without the case-by-case reasoning loses marks.

Ask the tutor: Want to try proving the product of four consecutive integers is always divisible by 24, using the same case-by-case style of argument?

Does this rule work for negative consecutive integers too, like -3, -2, -1?

Yes. The parity and multiple-of- $3$ spacing arguments do not depend on the numbers being positive — among any three consecutive integers, positive or negative, one is still even and one is still a multiple of $3$ . Check it: $(-3)\times(-2)\times(-1) = -6$ , and $-6 \div 6 = -1$ exactly, so the claim holds. The algebraic argument with $n, n+1, n+2$ covers negative starting values just as well, since nothing in the reasoning assumed $n$ was positive.

Is the product of four consecutive integers always divisible by 24?

Yes, and it follows the same style of reasoning extended one step further, though Class 8 only requires you to justify the $k=2$ and $k=3$ cases. The general pattern — the product of $k$ consecutive integers is always divisible by $k!$ — gives $4! = 24$ for four consecutive integers. Try $5\times6\times7\times8 = 1680$ , and $1680 \div 24 = 70$ exactly. This is stated here only as enrichment, not as something you need to prove at this level.

Why can't I just test a few examples and call the divisibility rule proven?

Because testing examples only shows the rule works for the numbers you tried — it cannot rule out some future starting value breaking the pattern. A genuine proof has to work for an arbitrary, unspecified $n$ , which is exactly what writing the integers as $n, n+1, n+2$ and reasoning about remainders achieves: the argument never uses a specific number, so it automatically covers every possible case at once, including numbers far too large to check by hand.

Products of consecutive integers

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads