A certain game is played with a total of 10 cards, numbered 1 through

For winning the game, player needs 2 cards whose product is 6. So, the only options are (1,6), (2,3),(3,2),(6,1).

All the cards other than 1,2,3 & 6 whose product is not 6 will not let the player win. So, 1,2,4,5,7,8,9,10.

IMO: D

I am a bit confused.

There are only 2 combinations that give the product = 6 --- (2,3) (1,6)

The winner can hold only one of the two pairs from above.

Then the other pair must be with the second player, and he also wins.

Bunuel : This means if a player holds cards with a product of 6 as well as other cards, he will not win.

Only then the answer can be 8.

Please clarify

I think the problem is poorly worded. however solve by considering each player has 10 cards.

OFA

Begin by finding the factors of 6. Factoring 6 produces 1, 6 and 2, 3. Since a player wins the game by having two cards in his or her hand whose product is 6, if a player has both the 1 and 6 cards or both the 2 and 3 cards in his or her hand, that player wins. Since 4, 5, 7, 8, 9, and 10 are not factors of 6, a player can have these six cards in his or her hand without winning. In addition, a player can have one each of the pairs of 1 and 6 or 2 and 3 in his or her hand without winning. For example, a player can have the hand containing the eight cards 1, 2, 4, 5, 7, 8, 9, and 10 without having any two of these cards have a product of 6. However, adding either of the two remaining cards 3 or 6 causes the hand to have two cards with a product of 6, thereby causing the player to win the game. Therefore, the maximum number of cards that a player can have in his or her hand without winning is eight. The correct answer is choice D.

1*2*4*5*7*8*9*10=201,600, which is multiple of 6.
1*2*4*5*7*8*10 = 22,400, which is not multiple of 6. So, correct answer should be 7.

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