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1) Suppose n is an arbitrary integer

(i) Show that $n(n + 1)$ is divisible by $2$

(ii) Show that $n(n + 1)(n + 2)$ is divisible by $3!$

attempt: Not sure if it's correct.

i) $n^2 + n = $even number

Two odd numbers added together must be an even number, so $n^2 + n$ will always be an even number, so it is divisible by $2$.

ii) note sure how to do.

Mauro ALLEGRANZA
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harold232
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3 Answers3

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$n(n+1)(n+2)$ is the product of three consecutive integers. Therefore at least one is divisible by $2$ and one by $3$. According to Euclid's lemma as $2, 3$ are coprimes, $n(n+1)(n+2)$ is divisible by $6$.

mathcounterexamples.net
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For the first part notice that in two consecutive integers one of them is odd and the other is even so the product $n(n+1)$ is even.

For the second one notice that in three consecutive integers one of them is a multiple of $3$ and at least one is multiple of $2$ so the product is multiple of $6$

Mohammad Riazi-Kermani
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HINT:

(i) Out of two consecutive integers one is always even. If one factor of a product is even then the product is even

(i) Out of three consecutive integers one is always divisible by $3$. If a number is divisible both by $2$ and $3$ then it is divisible by $2\cdot 3=6$. Use this and (i) to get the desired result

Hope this helped

Vinyl_cape_jawa
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