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Is the prime number p equal to 37 ? (1) p = n2 + 1, where n

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Is the prime number p equal to 37 ? (1) p = n2 + 1, where n [#permalink]

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18 Aug 2010, 07:08
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Is the prime number p equal to 37 ?

(1) p = n2 + 1, where n is an integer.

(2) p2 is greater than 200.
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Re: How to solve these kinds of questions [#permalink]

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18 Aug 2010, 15:05
zest4mba wrote:
Is the prime number p equal to 37 ?

(1) p = n2 + 1, where n is an integer.

(2) p2 is greater than 200.

it should be E

(1)

n=2 ==> 5 prime (No)
n=4 ==> n^2 +1 =17 prime (No)
n=6 ==> n^2 +1 =37 prime (yes)

2 p^2>200

17^2 >200

1 + 2 not suff

hence E
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Re: How to solve these kinds of questions [#permalink]

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18 Aug 2010, 20:33
I feel its E.

1) For n=1,2,3,4...6...infinity,w here n=6, p=37. But it p could have many values as there is no restriction on n.
2) Only info from here is that p > 15.

Combining 1 and 2, only info we get is p>15 so n>4 only, which is still insufficient.

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Re: How to solve these kinds of questions [#permalink]

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20 Aug 2010, 02:27
I guess C

1) simplify the equation a bit to sqrt(P-1) = n and plug in a couple of primes

n being an interger is true for more than one prime

n=4 => P=5, for n=6 => P=37, etc.

Insuff..

2) every prime greater than 17 when squared is more than 200

Insuff..

Now two together, solve for equation sqrt(p-1)=n.... what prime number (greater than 17) when subtracted by 1 and brought to its square root, results in a interger....... 37

Sqrt(37-1)=6

You can run a couple more prime numbers to check, 41, 43, none result in n being an interger!!

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Re: How to solve these kinds of questions [#permalink]

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27 Aug 2010, 04:37
I guess C

1) simplify the equation a bit to sqrt(P-1) = n and plug in a couple of primes

n being an interger is true for more than one prime

n=4 => P=5, for n=6 => P=37, etc.

Insuff..

2) every prime greater than 17 when squared is more than 200

Insuff..

Now two together, solve for equation sqrt(p-1)=n.... what prime number (greater than 17) when subtracted by 1 and brought to its square root, results in a interger....... 37

Sqrt(37-1)=6

You can run a couple more prime numbers to check, 41, 43, none result in n being an interger!!

In your explanation st 1 and st 2 and insufficient which is correct. Now when solving together we have two cases and just not one case:

Sqrt(37-1)=6 where both conditions are fulfilled ( 37^2 >200 and 37 = 6^2+1)
and

Sqrt(17-1)=4 where both conditions are fulfilled ( 17^2 >200 and 17 = 4^2+1)

Hence not sufficient.
Ans- E
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Re: How to solve these kinds of questions   [#permalink] 27 Aug 2010, 04:37
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