## Task B-3.4.

Tena and Ante are playing a game: first Tena says the number 0, then Ante says the number that is 1 greater. Then Tena says the number that is 2 greater than Ante's, and Ante says the number that is 3 greater than Tena's...

The game ends when someone says a number greater than 5050. What is that number? Who will win?
## Solution.

Let's see how Tena and Ante form the numbers:

| Tena | Ante |
| :--- | :--- |
| 0 | $0+\mathbf{1}=1$ |
| $1+\mathbf{2}=3$ | $3+\mathbf{3}=6$ |
| $6+\mathbf{4}=10$ | $10+\mathbf{5}=15$ |
| $15+\mathbf{6}=21$ | $21+\mathbf{7}=28$ |
| $28+\mathbf{8}=36$ | $\ldots$ |

The numbers are formed as follows:

$$
0, \quad 0+1, \quad 0+1+2, \quad 0+1+2+3, \quad 0+1+2+3+4, \quad \ldots
$$

So, in the $n$-th step, the number spoken will be $0+1+2+3+\cdots+(n-1)$.

This sum is equal to $\frac{(n-1) n}{2}$

and it must be $\frac{(n-1) n}{2}>5050$ or $(n-1) n>10100=100 \cdot 101$

i.e., $n>101$. Therefore, $n=102$

so Ante wins (since 102 is an even number).

The number he spoke is $\frac{102 \cdot 101}{2}=5151$.