Divisibility of an integer quantity

Number Theory Level 2

Regardless of what integer number n n you choose, the following quantities can always be divided by the same number. Which one?

n 3 + 6 n 2 + 11 n + 6 n^3+6n^2+11n+6

n 3 − 9 n 2 + 26 n − 24 n^3-9n^2+26n-24

n 3 − n n^3-n

9 6 4 8

Steve Gualtieri by Steve Gualtieri , 29, Italy

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2 solutions

Steve Gualtieri
Jan 29, 2018

When you decompose the quantities you get:

n 3 + 6 n 2 + 11 n + 6 = ( n + 1 ) ( n + 2 ) ( n + 3 ) n^3+6n^2+11n+6=(n+1)(n+2)(n+3)

n 3 − 9 n 2 + 36 n − 24 = ( n − 1 ) ( n − 2 ) ( n − 3 ) n^3-9n^2+36n-24=(n-1)(n-2)(n-3)

n 3 − n = ( n − 1 ) n ( n + 1 ) n^3-n=(n-1)n(n+1)

These are all products of three consecutive numbers, so it is guaranteed that at least one of them is even and exactly one is divisible by three, so that all of the products are always divisible by 6.

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I see that someone had answered while I was editing, since I had written down a wrong polynomial. I am sorry if this made him get the wrong answer.

Steve Gualtieri - 3 years, 4 months ago

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Don't worry I answered 6 (and then I realized that the middle polynomial that you changed n 3 + 3 n 2 − n − 3 = ( n − 1 ) ( n + 1 ) ( n + 3 ) n^3+3n^2-n-3 = (n-1)(n+1)(n+3) was not always divisible by 6 6 . Good problem.

Romain Bouchard - 3 years, 4 months ago
Siva Budaraju
Jan 29, 2018

Just put 2 into the third equation. The only option it is divisible by is 6.

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