Spain Mathematical Olympiad Problem 1

$\Large{\mathbf{1.\, Solution}}$

Let $x_1, \dots, x_{16}$ be a pucelana sequence.

Since such a sequence is constructed from consecutive odd numbers, we have $x_{i+1} = x_i + 2$ for $i = 1, \dots, 15$ . Or, more generally, $x_j = x_i + 2(j-i) \tag{1}$ where $i, j \in \mathbb{N}$ . It follows that the sum of the sequence is \begin{aligned} \sum_1^{16}{x_i} & = x_1 + (x_1 + 2(1)) + \dots + (x_1 + 2(15)) & \mbox{(by use of }(1)\mbox{)}\\ & = 16x_1 + 2\sum_1^{15}{r} \tag{2} \\ & = 16x_1 + 240 = n^3 \tag{3} \end{aligned} Now, we have been given that every $x_i$ is composed of exactly $3$ -digits. Thus the lower upper bounds for our cube number are $16(100) + 240 = 1840 > 12^3 \\ 16(999) + 240 = 16224 <\, 26^3$ respectively. This provides the constraint $12 < n < 26$ Let us rearrange $(3)$ to $x_1 = \frac{n^3}{16} - 15 \tag{4}$ We can observe that for $x_1$ to be an integer value we must have $16\, (= 2^4) \mid n^3 \iff 4 \mid n$ . Therefore our values for $n$ must be multiples of $4$ , giving $n = 16, 20, 24$ .

So our answer is $\fbox{3}$ .

$\Large{\mathbf{2.\, Generalisation}}$

There are multiple variables we can change in the given question. This section will generalise each of variables independently and, in some cases, simultaneously.

$\color{#FFFFFF}{Prepare yourselves...}$

$\large{\mathbf{2.1}\, d \mathbf{\mbox{-digit numbers}}}$

This is the most trivial generalisation. If the question were rephrased to ask the number of pucelana sequences with $d$ -digit numbers only, what would our answer be?

All that changes in our solution is the constraint on $n$ . For the lower and upper bounds we simply replace $100$ and $999$ with $10^{d-1}$ and $10^d-1$ (the smallest and largest numbers composed of $d$ -digits) respectively.

$\mathbf{Example\, 2.1.1}$ The same question for $4$ -digits would give lower and upper bounds of $16(1000) + 240 = 16240 > 25^3 \\ 16(9999) + 240 = 160224 <\, 55^3$ Thus our answer would be the number of multiples of $4$ between $25$ and $55$ , which is $7$ ( $28, 32, 36, 40, 44, 48, 52$ ).

$\color{#FFFFFF}{This next section was somewhat tedious to write}$

$\large{\mathbf{2.2}\, k\, \mathbf{\mbox{consecutive odd numbers}}}$

If, instead of $16$ , we were to define an increasing sequence with $k$ consecutive odd numbers whose sum is a perfect cube and ask the same question, what would our answer be?

Our formula $(1)$ and our constraints for $n$ will remain the same since we're still dealing with consecutive odd numbers of exactly $3$ -digits. The only thing to change would be our sum, which will now have $k$ terms. It is easy to see that equation $(2)$ will now become $kx_1 + 2\sum_1^{k-1}{r} = kx_1 + k(k-1) = n^3$ If we rearrange and divide by $k$ , akin to $(4)$ , we get $x_1 = \frac{n^3}{k} - (k-1) \tag{5}$

$\mathbf{Theorem\, 2.2.1}\, k \mid n^3 \iff p^{\lceil m/3 \rceil} \mid n \mbox{ for every prime factor } p \mbox{ of }$ $k \mbox{ where at most } p^m \mid k \mbox{ for some } m \in \mathbb{N}$

Proof. ( $\Rightarrow$ ) Let $k = pq$ where $p$ is a prime number and $q \in \mathbb{N}$ . If $p \nmid q$ then $k\, (=pq) \mid n^3 \implies p \mid n^3 \iff p \mid n$ . If $p \mid q$ then find $l \in \mathbb{N}$ such that $l = q/p^x$ and $q/p^{x+1} \not\in \mathbb{N}$ for some $x \in \mathbb{N}$ . Then at most $p^{x+1} \mid k$ so $k = p^{x+1}l$ and $p^{x+1} \mid n^3$ $\iff p^{\lceil (x+1)/3 \rceil} \mid n$ .

( $\Leftarrow$ ) Let $p_1^{\lceil m_1/3 \rceil}, p_2^{\lceil m_2/3 \rceil}, \dots, p_s^{\lceil m_s/3 \rceil} \mid n$ where $p_1, p_2, \dots, p_s$ are the prime factors of $k$ with at most $p_1^{m_1}, p_2^{m_2}, \dots, p_s^{m_s} \mid k$ . Thus $k = p_1^{m_1}p_2^{m_2} \cdots p_s^{m_s}$ . If $\frac{m_i}{3}$ is an integer for some $m_i$ then $\lceil \frac{m_i}{3} \rceil = \frac{m_i}{3} = x_i \in \mathbb{N}$ and thus $p_i^{m_i} \mid n^3$ . If however $\frac{m_i}{3}$ is not an integer then $\lceil \frac{m_i}{3} \rceil = \frac{m_i}{3} + c = x_i + c \in \mathbb{N}$ where $c = \lceil x_i \rceil - x_i$ and we have $p_i^{m_i + 3c} \mid n^3$ $\implies p_i^{m_i} \mid n^3$ . And so $k \mid n^3$ . $\square$

We can now use $\small{\mathbf{Theorem\, 2.2.1}}$ to find the number of integer solutions for $(5)$ .

$\mathbf{Example\, 2.2.2}$ Let us consider the number of sequences of $1458$ consecutive odd numbers whose sum is a cube number and with each term having exactly $4$ -digits only. It is simple to see that $1458 = 2 \times 729 = 2 \times 3^6$ And so, by $\small{\mathbf{Theorem\, 2.2.1}}$ , we require $n$ to be divisible by $2^{\lceil 1/3 \rceil} \times 3^{\lceil 6/3 \rceil} = 2 \times 3^2 = 18$ Since we already know ( $\small{\mathbf{Section\, 2.1}}$ ) that for $4$ -digit numbers the constraint on $n$ is $25 < n < 55$ we need to find all $n$ satisfying the inequality such that $18 \mid n$ . These are $36$ and $54$ and so our answer is $2$ .

$\color{#FFFFFF}{No point rewriting the theorem, right?}$

$\large{\mathbf{2.3}\, \mathbf{\mbox{ Sum equal to }} n^z}$

Finally we should consider what would happen if we required the sequence to be equal to $n^z$ for $z \in \mathbb{N}$ . In fact, it is nothing too new; we simply generalise $\small{\mathbf{Theorem\, 2.2.1}}$ as follows.

$\mathbf{Theorem\, 2.3.1}\, k \mid n^z \iff p^{\lceil m/z \rceil} \mid n \mbox{ for every prime factor } p \mbox{ of }$ $k \mbox{ where at most } p^m \mid k \mbox{ for some } m \in \mathbb{N}$

The proof is omitted (it is very similar to that of $\small{\mathbf{Theorem\, 2.2.1}}$ ).

So, now that we have done all of that, let's try some example questions.

$\Large{\mathbf{3.\, Example\, Questions}}$

$1$ . How many sequences are there of $32$ consecutive odd numbers with $2$ -digit numbers only, whose sum is of the form $n^4$ ?

$2$ . How many sequences are there of $30$ consecutive odd numbers with $3$ -digit numbers only, whose sum is of the form $n^2$ ?

Harder:

$3$ . What is the sum of the number of sequences of $3^{k+1}$ consecutive odd numbers with $k$ -digit numbers only, each with sum of the form $n^k$ for $k = 1, 2, 3, \dots$ .

$4$ . What is the sum of the number of sequences of $k^{k^2 - 2k + 1}$ consecutive odd numbers with $k$ -digit numbers only, each with sum of the form $n^k$ for $k = 1, 2, 3, \dots$ .