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# In a set of 11 consecutive integeres what is the smallest

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Director
Joined: 17 Dec 2005
Posts: 548
Location: Germany
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Kudos [?]: 16 [0], given: 0

In a set of 11 consecutive integeres what is the smallest [#permalink]  01 Jan 2006, 12:49
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In a set of 11 consecutive integeres what is the smallest integer ?
a) if x is the smallest integer then (x+72)^1/3 =4
b) if x is the smallest and y is the largest in the set then 16x^-2=y^-2
VP
Joined: 21 Sep 2003
Posts: 1065
Location: USA
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Kudos [?]: 40 [0], given: 0

D?

1) x = -8

2) 16/x^2-1/y^2 = 0
=> either 4/x-1/y =0 or 4/x+1/y =

Consider 4/x-1/y = 0
=> x = 4y ---(1)
Also since they are 11 consecutive integers y-x = 10 ---(2)
y-4y = 10
or y = -10/3 (NOT an integer) so x =4y cannot be TRUE

Hence 4/x+1/y = 0
or x = -4y
Using (1) y - (-4y) = 10 or y = 2
Hence x = -8
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Senior Manager
Joined: 11 Nov 2005
Posts: 328
Location: London
Followers: 1

Kudos [?]: 8 [0], given: 0

go for D.

From stat.1
(x+72)^1/3 =4
or, (x+72)= 4*4*4=64

so x= -8, sufficient

from stat.2, 16/x^2=1/y^2

if i put the value of x as -8, (just as picking a number not from stat 1, a short cut) then y=2, which satisfy x y are smallest and largest of 11 consequtive integers. Other values of x/y will not satisfy 16/x^2=1/y^2
and therefore from stat2. x=-8
Sufficient.
Intern
Joined: 30 Dec 2005
Posts: 1
Followers: 0

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

It is apparent that

a) if x is the smallest integer then (x+72)^1/3 =4
is sufficient because x can be evaluated. Then the remaining numbers can be dedced.

But
b) if x is the smallest and y is the largest in the set then 16x^-2=y^-2
can be resolved as follows:
(4^2)(x^-2)=y^-2
((4^2)(x^-2))^-1=(y^-2)^-1
(4^-2)(x^2)=y^2
(4^-1)x=y
x=4y

This is definitely not sufficient to solve the problem.

PostPosted: Sun Jan 01, 2006 12:49 pm Post subject: DS - consecuitve integers - find x Reply with quote
I've picked up this questions in the forum. The answer remained unclear that's why I post it again. I've no OA! Please show working.
Senior Manager
Joined: 14 Apr 2005
Posts: 418
Location: India, Chennai
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Kudos [?]: 6 [0], given: 0

giddi77 wrote:
D?

1) x = -8

2) 16/x^2-1/y^2 = 0
=> either 4/x-1/y =0 or 4/x+1/y =

Consider 4/x-1/y = 0
=> x = 4y ---(1)
Also since they are 11 consecutive integers y-x = 10 ---(2)
y-4y = 10
or y = -10/3 (NOT an integer) so x =4y cannot be TRUE

Hence 4/x+1/y = 0
or x = -4y
Using (1) y - (-4y) = 10 or y = 2
Hence x = -8

I agree. I also liked the way you solved this problem.
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