6. The sum of some 50 consecutive natural numbers is a perfect square. What are the two smallest possible sums of such 50 consecutive numbers?
First solution.

## Let

$n-24, n-23, \ldots, n-1, n, n+1, n+24, n+25$

be fifty consecutive natural numbers, $n \geq 25$.

2 POINTS

Then,

$n-24+n-23+\ldots+n-1+n+n+1+\ldots n+24+n+25=50 n+25, \quad 1$ POINT

i.e., $50 n+25=25(2 n+1). \quad 1$ POINT

Since 25 is a perfect square, then $2 n+1$ must also be a perfect square under the condition $n \geq 25$. 1 POINT

From $2 n+1 \geq 51$ and $2 n+1$ being an odd number, it follows that the smallest perfect square is 81. 2 POINTS

Thus, $2 n+1=81$ and the sum is $25(2 n+1)=25 \cdot 81=2025$. 1 POINT

Under the same conditions for $2 n+1$, the next largest perfect square is 121, 1 POINT

and the sum is $25(2 n+1)=25 \cdot 121=3025$. 1 POINT

The two smallest possible sums are 2025 and 3025.

TOTAL 10 POINTS

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