Shapovalovo $A$.

On the table were two decks, each with 36 cards. The first deck was shuffled and placed on top of the second. Then, for each card in the first deck, the number of cards between it and the same card in the second deck was counted (i.e., how many cards between the sevens of hearts, between the queens of spades, etc.). What is the sum of the 36 numbers obtained?
Let's number the cards from top to bottom in order. In the upper deck, the numbers are from 1 to 36, and in the lower deck, the numbers are from 37 to 72. Let the $i$-th card of the upper deck coincide with the $k_{i}$-th card of the lower deck ($i=1,2, \ldots, 36$). Between them lies $k_{i}-i-1$ cards, so the desired sum $S=\left(k_{1}-1-1\right)+\left(k_{2}-2-1\right)+\ldots+\left(k_{36}-36-1\right)$. By rearranging the terms $k_{1}, k_{2}, \ldots, k_{36}$ in ascending order, we do not change the sum. Therefore, $S=(37-1-1)+(38-2-1)+\ldots+(72-36-1)=36 \cdot 35=1260$.

## Answer

1260. Send a comment

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Author: Berpovo S.l.

Some natural numbers are marked. It is known that on each segment of the number line of length 1999, there is a marked number.

Prove that there will be a pair of marked numbers, one of which divides the other.

## Solution

Consider segments of the natural number sequence of length 1999. All of them differ by a shift. We will call the points, or positions, on two such segments corresponding if they coincide when one segment is shifted to the other.

Assume that no marked number divides another.

Consider the segment $[1,1999]$. By the condition, it contains a marked number, say $x_{1}$. Shift this segment by $x_{1}$. At the position corresponding to the number $x_{1}$, there cannot be a marked number (since it is divisible by $x_{1}$), but there is some other marked number $x_{2}$ in this segment. Shift the new segment by $x_{1} x_{2}$. Then at the positions corresponding to the marked numbers $x_{1}$ and $x_{2}$, there cannot be marked numbers (since these numbers are divisible by $x_{1}$ and $x_{2}$, respectively), but there is some marked number $x_{3}$ in this segment.

Shift the new segment by $x_{1} x_{2} x_{3}$, find a new number $x_{4}$, and so on. At step $t$, we perform a shift by $x_{1} \ldots x_{t}$ and obtain a segment with $t$ prohibitions. At step 1999, we will get that all positions are prohibited, which contradicts the condition.