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

Question

A certain game is played with a total of 10 cards, numbered 1 through 10. A player wins if his or her hand contains two cards with numbers whose product is 6. What is the maximum number of cards that a player can have in his or her hand without winning?

A) Five
B) Six
C) Seven
D) Eight
E) Nine

A) Five
B) Six
C) Seven
D) Eight 
E) Nine

sumi747 · Answer

Factors pairs of 6= 2*3,1*6 
So the rest of the 6cards can be held without the product of 2 cards being 6.
Now if we are also holding 2 and 1, we can be holding 8 cards without the product of 2 cards as 6.
Hence we can be holding 8cards without making a product of 2 cards as 6, but we have to make sure that when we are holding 1 we cannot be holding 6 and when we are holding 2 we cannot be holding 3.
IMO D

Posted from my mobile device

sumisachan · Answer

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

AkshdeepS · Answer

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

sach24x7 · Answer

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

buan15 · Answer

Don't really understand this question.
If the winner has any of the pair   or   then also the remaining cards can give multiple of 6 for e.g.   ?
Don't we need to eliminate 9 as well? Then I get 7 as the answer. 
 Bunuel  Can you please explain?

Thanks !

Kaushik786 · Answer

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.

Kushagra041 · Answer

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

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

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