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The $n^{th}$ term of a sequence is given by the formula $n^3 +
11n$. Find the first four terms of the sequence given by this
formula and the first term of the sequence which is bigger than one
million. Prove that all terms of the sequence are divisible by
$6$.
Congratulations to Julia Collins, age 17, Langley Park School for
Girls, Bromley; Kookhyun Lee; Yatir Halevi age 17; Sim S K age 14;
Ang Zhi Ping age 16 for your splendid solutions.
This is Kookhyun Lee's solution: First term: $12$, second term:
$30$, third term: $60$, fourth term: $108$. The $100$th term is
$100^3 +1100 = 1001100$. The $99$th term is $970299 + 1089 =
971388$ so the first term bigger than a million is $1001100$ when
$n=100.$
Proof that all the terms are divisible by $6$.
$$n^3 + 11n = n^3 + 12n - n = n(n^2-1) + 12n$$
This must be a multiple of 6 because $n(n^2-1)$ can be written as
$(n-1)\times n \times (n+1)$. Any multiple of three consecutive
integers is a multiple of $6$ because it contains a multiple of two
(an even number) and a multiple of three.
The following solution, uses a different method. It arrived early
in the morning on the first day that the question was published
from Yatir Halevi, age 17.5, Maccabim and Reut High-School,
Israel.
We have a sequence given by the formula $n^3+11n.$ We have to find
the first value of $n$ for which $n^3+11n> 10^6$ and we could
use
http://www.sosmath.com/algebra/factor/fac11/fac11.html
The second way is a much nicer one. We notice that $100^3$ is
$10^6$, so we know for $n=100$ that $n^3+11n$ is bigger than
$10^6$, so we check $n=99$ and we get: $971388$ which is smaller
than $10^6$. So we have the answer: $n=100$ is the first $n$ for
which $n^3+11n$ is bigger than $10^6$.
The next thing we have to prove is that $n^3+11n$ is always
divisible by $6$. This we will prove by using modular arithmetic.
We will use modulus $6$. For each $n$, we can have a residue of
either: $0$, $1$, $2$, $3$, $4$ or $5$. For $n^3$ we get the
following residues:
$0$, $1$, $2$, $3$, $4$, $5$ respectively (to $n$). For $11n$ we
get the following residues: $0$, $5$, $4$, $3$, $2$, $1$
respectively (to $n$).
Combining $n^3$ and $11n$ (respectively) we get a $0$ residue,
because: $0+0=0$ (mod $6$), $1+5=6=0$ (mod $6$), $2+4=6=0$ (mod
$6$), $3+3=6=0$ (mod $6$), $4+2=6=0$ (mod $6$), $5+1=6=0$ (mod
$6$). This means that we get a zero residue when dividing by $6$,
or in other words, $(n^3+11n)$ is a multiple of $6$ or $6$ divides
$n^3+11n$.