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# each player scores either 2 points or 5 points. If n players

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Director
Joined: 12 Jun 2006
Posts: 531

Kudos [?]: 166 [0], given: 1

each player scores either 2 points or 5 points. If n players [#permalink]

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21 Oct 2006, 17:03
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

each player scores either 2 points or 5 points. If n players score 2 points and m players socre 5 points, and the total number of points scored is 50, what is the least possible positive difference between n and m?

(A) 1
(B) 3
(C) 5
(D) 7
(E) 9

Kudos [?]: 166 [0], given: 1

Director
Joined: 06 Sep 2006
Posts: 734

Kudos [?]: 49 [0], given: 0

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21 Oct 2006, 21:41
n*2 + m*5 = 50

m has to be even since it multplies by 5

so n,m = 10,6 result = 10 - 6 = 4
or n,m = 5,8 result = 3
or n,m = 0,10
or n,m = 15,4
I believe B is the answer.

Kudos [?]: 49 [0], given: 0

SVP
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 169 [0], given: 0

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22 Oct 2006, 02:55
(B) as well, this is another approch... On G day, It's better to pick numbers

n*2 + m*5 = 50

Let k represent the difference between n and m.

k=n-m
<=> n = m+k

min(|k|) = ?

n*2 + m*5 = 50
<=> 2*m+2*k + m*5 = 50
<=> k = (50-7*m)/2
=> |k| = |50-7*m|/2

After that, one way is:

If m = 0, |k| = 25
If m = 2, |k| = 18
If m = 4, |k| = 11
If m = 6, |k| = 4
If m = 8, |k| = 3

or

another way is:

f(x) = |50-7*x|/2. The minimum of an absolute value is 0. In another words, we must search x such that f(x) = 0.

f(x)=0
<=> 50-7*x = 0
<=> x=50/7

min(|k|) implies that m has to be the nearest integer of 50/7 that is divisible by 2. Thus, m = 8 and min(|k|) = |50-7*8|/2 = 6/2 = 3

Kudos [?]: 169 [0], given: 0

22 Oct 2006, 02:55
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# each player scores either 2 points or 5 points. If n players

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