The nth term of a sequence is given by the formula n³ + 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 and Andre Lazanu age 12 for your splendid solutions.

This is Kookhyun Lee's solution: First term: 12, second term: 30, third term: 60, fourth term: 108. The 100th term is 1003 + 1100 = 1001100. The 99th term is 970299 + 1089 = 97138 so the first term bigger than a million is 1001100 when n=100.

Proof that all the terms are divisible by 6.

n3 + 11n = n3 + 12n − n = n(n2−1) + 12n

This must be a multiple of 6 because n(n²−1) can be written as (n−1)×n ×(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 n³+11n. We have to find the first value of n for which n³+11n > 10⁶ and we could use http://www.sosmath.com/algebra/factor/fac11/fac11.html

The second way is a much nicer one. We notice that 100³ is 10⁶, so we know for n=100 that n³+11n is bigger than 10⁶, so we check n=99 and we get: 970398 which is smaller than 10⁶. So we have the answer: n=100 is the first n for which n³+11n is bigger than 10⁶.

The next thing we have to prove is that n³+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 n³ 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 n³ 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, (n³+11n) is a multiple of 6 or 6 divides n³+11n.