Míša is exploring numbers that can be expressed as the sum of at least two consecutive natural numbers. She is particularly interested in numbers that can be expressed in multiple ways (for example, $18=5+6+7=3+4+5+6$). She calls numbers that can be expressed in at least three different ways grand.

Find at least three of Míša's grand numbers.

(V. Hucíková)

Hint. What results can be obtained by the sum of two, three, etc., consecutive natural numbers?
Two consecutive numbers give sums

$$
1+2=3, \quad 2+3=5, \quad 3+4=7, \quad \ldots
$$

The addends are increased by 1 sequentially, so the sums increase by 2 sequentially.

Three consecutive numbers give sums

$$
1+2+3=6, \quad 2+3+4=9, \quad 3+4+5=12
$$

The addends are increased by 1 sequentially, so the sums increase by 3 sequentially.

Further, we find that the smallest sum of four consecutive numbers is $1+2+3+4=$ $=6+4=10$ and the next possible sums are $14,18,22, \ldots$

Similar considerations give us the following overview of sums of several consecutive numbers:

| sums of two | 3 | 5 | 7 | 9 | 11 | 13 | 15 | 17 | 19 | 21 | 23 | $\ldots$ |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :--- |
| sums of three | 6 | 9 | 12 | 15 | 18 | 21 | 24 | 27 | 30 | 33 | $\ldots$ |  |
| sums of four | 10 | 14 | 18 | 22 | 26 | 30 | 34 | 38 | 42 | $\ldots$ |  |  |
| sums of five | 15 | 20 | 25 | 30 | 35 | 40 | 45 | $\ldots$ |  |  |  |  |
| sums of six | 21 | 27 | 33 | 39 | 45 | $\ldots$ |  |  |  |  |  |  |

The sought grand numbers are those numbers that belong to at least three of the above-mentioned groups. The three smallest grand numbers (and their respective decompositions) are:

$$
\begin{gathered}
15=7+8=4+5+6=1+2+3+4+5 \\
21=10+11=6+7+8=1+2+3+4+5+6 \\
27=13+14=8+9+10=2+3+4+5+6+7
\end{gathered}
$$

Note. There are multiple ways to find grand numbers. For example, for any six consecutive numbers, the sum of the first and sixth number is the same as the sum of the second and fifth, and that is the same as the sum of the third and fourth; this sum is odd and we denote it as $a$. The sum of all six numbers is then equal to $3a$, which is a number that can be expressed as the sum of three consecutive numbers $a-1, a, a+1$. Since $a$ is an odd number, $3a$ is also odd and any such number is the sum of two consecutive numbers; in the current notation $\frac{1}{2}(3a+1)$ and $\frac{1}{2}(3a-1)$.

It also holds that the sum of an odd number of consecutive numbers is always a multiple of that number. All these (and other interesting) insights can be successfully combined to find more grand numbers. From the above, it follows that there is an infinite number of grand numbers.