Another Inequality

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Another Inequality

by knight247 » Tue Aug 23, 2011 3:16 am
If x1,x2,x3....xn is a sequence of positive integers such that xn+2=xn*xn+1. What is the least possible value of x1 if |x3-x2|=10?
(A)3
(B)5
(C)4
(D)1
(E)2

The OA says A But I'm getting a value of 2

Here is how i did it
|x3-x2|=10
|x2*x1-x2|=10
|x2||x1-1|=10

Since both x2 and x1 are positive
the options are
x2=5 and x1-1=2 so x1=3
or
x2=2 and x1-1=5 so x1=6
Hence, x1=3 as it is the minimum value

But, can't we also consider another option as follows?
x2=10 and x1-1=1 so x1=2
or
x2=1 and x1-1=10 so x1=11

Making x1=2 and hence making the answer E. Opinions?
Source: — Problem Solving |

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by force5 » Tue Aug 23, 2011 5:13 am
But, can't we also consider another option as follows?
x2=10 and x1-1=1 so x1=2
or
x2=1 and x1-1=10 so x1=11
im really sorry but can you please explain how are u considering this a contender? somehow i'm not able to follow what you have written

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by knight247 » Tue Aug 23, 2011 5:17 am
At force5
If
X*Y=10
Possible options are
X=5 Y=2 or vice versa

Or
X=10 and Y=1 or vice versa.

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by force5 » Tue Aug 23, 2011 6:10 am
ok so this is what you are saying

|x1*x2-x2|= 10

x2*|x1-1|=10

x2 can be 10 and x1 can be 2 ---- x3 = 20 which gives the value of x3-x2 as 10
or
x2 can be 1 and x1= 11 ---- x3 = 11 which gives value of x3-x2 as 10

hence lowest x1 we got till now is 2

Other part you have already done.....

So you are correct.... the lowest value has to be 2.