6. On 52 cards, the numbers from 1 to 52 are written. It is considered that 1 is older than 52, and in all other pairs, the card with the larger number is older. The cards are lying on the table in a random order face down. In one question, Petya can find out which card is older in any pair. Can Petya guarantee to find the card with 52 in no more than 64 questions?
6. Answer: it can. We will divide the cards into 13 quartets. In each quartet, we will make the following comparisons: by dividing the quartet into pairs, we will determine the higher cards in the pairs (let $\mathrm{a}>\mathrm{b}$ and $c>d$), then we will determine the higher of the two higher cards (let $a>c$). Card $a$ is higher than two cards, so $\mathrm{a} \neq 1$. Therefore, $\mathrm{b} \neq 52$ and $\mathrm{c} \neq 52$. Compare $a$ and $d$. If $a>d$, then $d \neq 52$, so the number $a$ is the highest in the quartet. Then we move on to the next quartet. If this situation repeats in all quartets, then in $4 \cdot 13=52$ questions we will eliminate 39 cards that are not 52, and among the 13 remaining cards, there will be no card 1. By comparing the remaining cards one by one and discarding the lower one, we will find the card 52 in another 12 questions. Suppose, however, that in some quartet during the fourth comparison it turned out that $\mathrm{a}<c>d>a$. It is clear that this is only possible if the quartet contains the numbers 1 and 52: $\mathrm{c}=1, \mathrm{~d}=52$ or $\mathrm{d}=1, \mathrm{a}=52$. Compare $\mathrm{b}$ and $\mathrm{d}$ ( $b \neq 52$ ). If $b>d$, then $a=52$, and if $b<d$, then $d=52$. Thus, we found the card 52 in no more than 53 questions.

Criteria: An algorithm for finding the required card is presented and justified, which always works - 7 points. In other cases - 0 points. Insufficient justification - minus $1-2$ points.