Shapovalov A.V.

Thieves Hapok and Glazok are dividing a pile of 100 coins. Hapok grabs a handful of coins from the pile, and Glazok, looking at the handful, decides who of the two will get it. This continues until one of them receives nine handfuls, after which the other takes all the remaining coins (the division may also end if the coins are divided before anyone gets nine handfuls). Hapok can grab any number of coins in a handful. What is the maximum number of coins he can guarantee himself regardless of Glazok's actions?
Here is Hapka's strategy, which will ensure him no less than 46 coins: take six coins each time (as long as it is possible). The pile contains 16 such full handfuls and four more coins. With this strategy, Hapka (regardless of how Glazok acts) has only two possibilities:

a) at some point, Glazok will have nine full handfuls, the division will end at this point, and the remaining 46 coins will be with Hapka

b) Glazok takes no more than eight full handfuls for himself, which means that at some point Hapka will receive eight such handfuls; at this moment, he will have no less than 48 coins.

Glazok's strategy, which will ensure him no less than 54 coins (which means that Hapka cannot guarantee himself more than 46 coins): to give Hapka the handfuls with fewer than six coins (small), and take the rest (large) for himself. With this strategy, Glazok (regardless of how Hapka acts) also has two possibilities:

a) Glazok will take nine large handfuls for himself, which means he will have no less than 54 coins;

b) Hapka will receive nine or fewer (if the division ends "prematurely") small handfuls, which means no more than 45 coins.

## Answer

46 coins.

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