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eyal
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A person has 5 cards: A,2,4,6,8 He can drop 1 to 5 cards 24 Mar 2004, 06:13
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A person has 5 cards: A,2,4,6,8
He can drop 1 to 5 cards down, and the sum of the cards will be added (say he drops 3 cards: 2,3,6, then the sum is 11)

How many different sums can he get?

A 15
B 21
C 22
D 20
E 5



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A person has 5 cards: A,2,4,6,8 He can drop 1 to 5 cards 24 Mar 2004, 07:47
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Are you sure the "4" in the card in hand isn't supposed to be a "3"? Otherwise, the answer should be 10
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A person has 5 cards: A,2,4,6,8 He can drop 1 to 5 cards Updated on: 24 Mar 2004, 08:43
Originally posted by BG on 24 Mar 2004, 08:39.
Last edited by BG on 24 Mar 2004, 08:43, edited 1 time in total.
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1 card-2,4,6,8-4 options, 2 cards-2+4,2+6,2+8,4+6,4+8,6+8-another 3 different sums (10,12,14), 3 cards-2+4+6,2+4+8,4+6+8-only 1 different sum( 18), 4 cards-2+4+6+8-another different sum (20) or 4+3+1+1=9.In total 9 different sums.
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A person has 5 cards: A,2,4,6,8 He can drop 1 to 5 cards 24 Mar 2004, 08:39
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eyal
A person has 5 cards: A,2,4,6,8
He can drop 1 to 5 cards down, and the sum of the cards will be added (say he drops 3 cards: 2,3,6, then the sum is 11)

How many different sums can he get?

A 15
B 21
C 22
D 20
E 5
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What is 'A' on the first card here?
I am missing something.............
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A person has 5 cards: A,2,4,6,8 He can drop 1 to 5 cards 24 Mar 2004, 08:44
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BG
1 card-2,4,6,8-4 options, 2 cards-2+4,2+6,2+8,4+6,4+8,6+8-another 3 different sums (10,12,14), 3 cards-2+4+6,2+4+8,4+6+8-only 1 different sum( 18), 4 cards-2+4+6+8-another different sum (20) or 4+3+1+1=9.In total 9 different sums.
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You forgot 16 = 8+6+2 so total should be 10 diff. sums
However, does A stand for ace=1?
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A person has 5 cards: A,2,4,6,8 He can drop 1 to 5 cards 24 Mar 2004, 08:44
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guess A stands for an ACE but not sure and do not know if we count it or not :?
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A person has 5 cards: A,2,4,6,8 He can drop 1 to 5 cards 24 Mar 2004, 09:00
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If A stands for ace=1 then answer should be (C) 22. All numbers from 0 to 21 inclusively. If he drops all 5 cards sum is 0. That was a trap :)
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A person has 5 cards: A,2,4,6,8 He can drop 1 to 5 cards 24 Mar 2004, 10:12
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Paul
If A stands for ace=1 then answer should be (C) 22. All numbers from 0 to 21 inclusively. If he drops all 5 cards sum is 0. That was a trap :)
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paul: Could you crarify as to how one can get a sum of ZERO.
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A person has 5 cards: A,2,4,6,8 He can drop 1 to 5 cards 24 Mar 2004, 10:14
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Paul
If A stands for ace=1 then answer should be (C) 22. All numbers from 0 to 21 inclusively. If he drops all 5 cards sum is 0. That was a trap :)

paul: Could you crarify as to how one can get a sum of ZERO.
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I refer you to the problem: "He can drop 1 to 5 cards down". If he slaps 5 cards down, then the sum in his hand is 0. Makes sense?
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A person has 5 cards: A,2,4,6,8 He can drop 1 to 5 cards 24 Mar 2004, 10:17
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Paul

I refer you to the problem: "He can drop 1 to 5 cards down". If he slaps 5 cards down, then the sum in his hand is 0. Makes sense?
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Is't the question asking for sum of the cards dropped? Am I missing something?
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A person has 5 cards: A,2,4,6,8 He can drop 1 to 5 cards 24 Mar 2004, 10:26
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kpadma
Paul

I refer you to the problem: "He can drop 1 to 5 cards down". If he slaps 5 cards down, then the sum in his hand is 0. Makes sense?

Is't the question asking for sum of the cards dropped? Am I missing something?
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:cry: You're right. I misread this question again. Yes, if it's sum of cards dropped, then 0 cannot be the sum since the min. number of cards dropped is 1... (B) 21 would be answer then
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A person has 5 cards: A,2,4,6,8 He can drop 1 to 5 cards 25 Mar 2004, 00:31
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By using 1,2,4,8 we can reach a sum equal to 15 (like binary counting)

By adding 6, we can reach numbers higher than 15 that we couldn't have reached before:
16 (8 + 2 + 6 for ex.)
17
18
19
20
21 (15+6)
--->21 (B)

0 is not a valid sum since the player has to drop at least one card down
A is a 1. :wink:



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