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a s equal to 100 multiplied by the largest integer that is 01 Sep 2006, 15:07
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€(n) is equal to 100 multiplied by the largest integer that is smaller than (n+50)/100. If a,b and c are distinct positive integers, is €(a+b+c)> €(a)+€(b)+€(c)?

(1) a+b+c<150
(2) abc < 100



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a s equal to 100 multiplied by the largest integer that is 02 Sep 2006, 09:36
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I believe the answer is E

From 1 a+b+c = 6(1,2,3) or a+b+c=75 (15,20,40)

If we try the above values we get €(a+b+c)> €(a)+€(b)+€(c) or €(a+b+c)= €(a)+€(b)+€(c)

INSUFF

From 2, abc < 100 i.e. abc can be (1,2,3) or (1,2,49) and again for the 2 sets we see €(a+b+c)= €(a)+€(b)+€(c) or €(a+b+c)> €(a)+€(b)+€(c)

INSUFF

Combined is also clearly INSUFF...
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a s equal to 100 multiplied by the largest integer that is 02 Sep 2006, 10:09
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Wouldn't the question boil down to :
E(a+b+c)=((a+b+c) + 50) and
E(a)+E(b)+E(c)=(a+50)+ ... = ((a+b+c) + 150) ?
Correct me if I'm wrong ...
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a s equal to 100 multiplied by the largest integer that is 06 Sep 2006, 15:21
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Writing this in a simpler way:-

€(n) = 100 f(n)....where f(n) is as follows:-

n=1-49 f(n) = 0
n=50-149 f(n) = 1
n=150-249 f(n) = 2

so we get:

n=1-49 €(n) = 0.100 = 0
n=50-149 €(n) = 1.100 = 100
n=150-249 €(n) = 2.100 = 200

From 1:
a+b+c €(a+b+c)
consider a=100, b=25, c=24
€(a)+€(b)+€(c) =100 = €(a+b+c)
consider a=25, b=26, c=27
€(a)+€(b)+€(c) =0 =50
Then
€(a)+€(b)+€(c) = 0.
also, lets try to find max of €(a+b+c) to see if it can be >0. For this a+b+c must be >=50.
For a=48, b=2, c=1
€(a+b+c) = 100 > €(a)+€(b)+€(c)
For a=1,b =2, c=3
€(a+b+c) = 100 = €(a)+€(b)+€(c)

Combining 1 and 2 does not make a difference, hence E



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