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 $n^3+11n$.