Spain Mathematical Olympiad Problem 3

It's really very easy. Just see that -

$111*a = \frac{n(n+1)}{2}$ , such that $a = [1,9]$ , which I think is the immediate reaction to the problem.

Then we know that $111 = 37*3$ , and that $111|n(n+1)$

Immediately, the minimum solution comes as $n+1 = 37$ because if $n = 37$ , then $n+1$ is not a multiple of $3$ .

And hence $666$ becomes the answer.

But what about the case above $37$ ? Well, it will be a multiple of $111$ only. The minimum of the immediate multiple will be for $n= 37*2$ and $n+1 = 25*3$ But here it comes out to be $111*25$ but wait $a = [1,9]$ and hence there are no more solutions(as if the 'minimum' of an increasing sequence doesn't work, how can the ones greater than it work?)

Hence, $\lfloor \frac{N}{3} \rfloor = 666$ . And maximum occurs obviously at $N = 666*3+2 = 2000$

$(i)$ and $(ii)$ imply that

$a\times 111 = \dfrac{n(n+1)}{2}$

for some $n \in \mathbb N$ and some $a \in \{1,2,3,4,5,6,7,8,9 \}$ .

Hence

$n(n+1) = 2\times a \times 111 = 2 \times a \times 3 \times 37$

$37$ is prime so it divides either $n+1$ or $n$ (and not both, couse two consecutive integers are prime).

This means that $n+1$ or $n$ is a multiple of $37$ . If this multiple was greater that $37$ itself, we would have it is at least $74$ . And hence we would get

$n(n+1) \geq 74*73 \Rightarrow a \times 2 \times 3 \times 37 \geq 74 \times 73 \Rightarrow 3a \geq 73$

against the fact that $a \leq 9$ .

So this multiple IS $37$ , and this means $n = 37$ or $n = 36$ .

If $n = 37$ we would get

$37\times38 = 2 \times a \times 3 \times 37 \Rightarrow 38 = 2 \times a \times 3$

that is impossible because $3$ doesn't divide $38$ .

The only chance is so $n= 36$ and this lead to $a = 6$ .

Trivially the greatest $N$ such that

$\left[\dfrac{N}{3}\right] = 666$

is $2000$ .

It is indeed problem 4 in year 2000.

Sum consecutive numbers $1+2+\ldots+k$ until you get over $1000$ . Indeed, you have to do it until $k=46$ , and get the (only) solution for $k=36$ . Of course this is a really bad solution, but it will not take more than 20 minutes, which is fine in an olympiad.

What about the problem with 4 or more equal digits?