Standard +0.5 This is a standard divisibility proof requiring students to consider cases modulo 2 and 3, showing one factor must be even and one divisible by 3. While it requires proof technique and case analysis beyond routine calculation, it's a well-known textbook problem with a straightforward approach that most prepared students can execute.
At least one of the numbers is even, and so the product is a multiple of 2.
M1
Divisibility by 2
If \(n\) is a multiple of 3 then so is the product.
M1
Divisibility by 3
If \(n = 3k+1\) then \(n+2\) is a multiple of 3
If \(n = 3k+2\) then \(n+1\) is a multiple of 3.
\(n\) must have one of the forms \(3k\), \(3k+1\) or \(3k+2\). Therefore whichever it is one of the three numbers is a multiple of 3 and so the product is a multiple of 3.
E1
Conclusion
Since it is also a multiple of 2 it is a multiple of 6.
4
Call the numbers $n$, $n+1$ and $n+2$ | B1 | Algebra
At least one of the numbers is even, and so the product is a multiple of 2. | M1 | Divisibility by 2
If $n$ is a multiple of 3 then so is the product. | M1 | Divisibility by 3
If $n = 3k+1$ then $n+2$ is a multiple of 3 | |
If $n = 3k+2$ then $n+1$ is a multiple of 3. | |
$n$ must have one of the forms $3k$, $3k+1$ or $3k+2$. Therefore whichever it is one of the three numbers is a multiple of 3 and so the product is a multiple of 3. | E1 | Conclusion
Since it is also a multiple of 2 it is a multiple of 6. | |
| 4 |