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Nice DS problem

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by The Iceman » Thu Dec 06, 2012 1:36 am
Let p(x) = x^2 + 40. Then for any two positive integers i and j where i > j, is p(i) + p(j) a composite number?

(I) p(i) - p(j) is not a composite number

(II) p(2i) + p(2j) is a composite number
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Source: — Data Sufficiency |
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by nisagl750 » Thu Dec 06, 2012 5:05 am
The Iceman wrote:Let p(x) = x^2 + 40. Then for any two positive integers i and j where i > j, is p(i) + p(j) a composite number?

(I) p(i) - p(j) is not a composite number

(II) p(2i) + p(2j) is a composite number
Statement1:
Lets take different values of i & j
i=3, j=2
p(i)-p(j) = 5 which is a Prime number (Not composite)
p(i)+p(j) = 93 Composite

i=5, j=2
p(i)-p(j) = 61 which is a Prime number (Not composite)
p(i)+p(j) = 109 which is a Prime number (Not composite)
Insufficient

Statement2:
Lets take different values of i & j
i=3, j=2
p(2i)+p(2j) = 132 which is composite
p(i)+p(j) = 93 Composite

i=5, j=2
p(2i)+p(2j) = 196 which is composite
p(i)+p(j) = 109 which is a Prime number (Not composite)
Insufficient

Statements 1 & 2 Together:
Take same examples
i=3, j=2 & i=5, j=2

Insufficent

Ans E
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by The Iceman » Thu Dec 06, 2012 8:49 am
nisagl750 wrote:
i=5, j=2
p(i)-p(j) = 61 which is a Prime number (Not composite)
p(i)+p(j) = 109 which is a Prime number (Not composite)
Insufficient

Calculation mistake highlighted in red. Please try again.
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by nisagl750 » Thu Dec 06, 2012 9:09 am
The Iceman wrote:
Calculation mistake highlighted in red. Please try again.
I have corrected. Hope It makes sense now.....

Statement1:
Lets take different values of i & j
i=3, j=2
p(i)-p(j) = 5 which is a Prime number (Not composite)
p(i)+p(j) = 93 Composite

i=4, j=3
p(i)-p(j) = 7 which is a Prime number (Not composite)
p(i)+p(j) = 105 which is composite

i=6, i=5
p(i)-p(j) = 11 which is a Prime number (Not composite)
p(i)+p(j) = 141 which is composite

sufficient

Statement2:
Lets take different values of i & j
i=3, j=2
p(2i)+p(2j) = 132 which is composite
p(i)+p(j) = 93 Composite

i=5, j=2
p(2i)+p(2j) = 196 which is composite
p(i)+p(j) = 109 which is a Prime number (Not composite)
Insufficient

Statements 1 & 2 Together:
Take same examples
i=3, j=2 & i=5, j=2

Insufficent

Ans A
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by The Iceman » Thu Dec 06, 2012 10:04 am
nisagl750 wrote: i=5, j=2
p(2i)+p(2j) = 196 which is composite
p(i)+p(j) = 109 which is a Prime number (Not composite)
Insufficient

Statements 1 & 2 Together:
Take same examples
i=3, j=2 & i=5, j=2

Insufficent

Ans A
Bingo!

Statement I can be validated using a generic process.

p(i) - p(j) is not a composite number
=>i^2-j^2 is a prime as i ,j are positive integers and i >j, ( i^2-j^2) can't be 1
=>(i+j)(i-j)= prime
so i-j=1
let p be the prime so i=(p+1)/2 and j=(p-1)/2
clearly p is not 2, hence p is odd
p(i) + p(j)=80 +(p^2+1)/2
for p =3, p(i) + p(j)=85, which is not a prime

Also p^2=6k+1 ( for p>3 )

=>p(i) + p(j)=80 +(p^2+1)/2= 80+(6k+2)/2 = 81+3k = 3(27+k), which is not a prime => can be answered using I
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