[ [motion problem ]

In a cycling competition on a circular track, Vasya, Petya, and Kolya participated, starting simultaneously. Vasya completed each lap two seconds faster than Petya, and Petya completed each lap three seconds faster than Kolya. When Vasya finished the distance, Petya had one lap left to complete, and Kolya had two laps left. How many laps did the distance consist of?

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Let the distance be $n$ laps. Then, over the entire distance, Vasya lapped Petya by $2n$ seconds (during which Petya completed the last lap), and Kolya by $5n$ seconds (during which Kolya completed 2 laps). This means Kolya took $2.5n$ seconds per lap, which is $0.5n$ seconds slower than Petya. According to the condition, $0.5n = 3$.

## Answer

6 laps.

Petya found the sum of all odd divisors of some even number (including 1), and Vasya found the sum of all even divisors of the same number (including the number itself). Can the product of the two found numbers be a perfect square?

## Solution

Let $2^n$ be the highest power of two that divides the given number. If Petya obtained a set of its odd divisors $a_1, a_2, \ldots, a_m$, then in Vasya's set of even divisors, there should be all numbers that are obtained by multiplying each odd divisor by each power of two from $2^1$ to $2^n$. Thus, all numbers in Vasya's set have the form $2^k a_i$, where $1 \leq k \leq n, 1 \leq i \leq m$.

Let the sum of Petya's numbers be $A$. Then the sum of Vasya's numbers is $A \cdot (2 + 2^2 + 2^3 + \ldots + 2^n)$.

Therefore, the product of Petya's and Vasya's sums is $A^2 \cdot (2 + 2^2 + 2^3 + \ldots + 2^n)$. For this number to be a perfect square, it is necessary that the expression in parentheses be a perfect square. However, the sum of the powers of two, when divided by 4, gives a remainder of 2, meaning it is even but not divisible by 4.

## Answer

No, it cannot.