Is the positive integer p prime? (1) p = n^2 - n + 41, where n is an

Question

Is the positive integer p prime?

(1) p = n^2 - n + 41, where n is an integer
(2) p < 43

chetan2u · Answer

(1) p = n^2 - n + 41, where n is an integer
 p = n^2 - n + 41=n(n-1)+41 ...
Clearly n as 0 or 1 will give p=41, which is prime.
But surely when n or n-1 is 41, p will not be a prime..
p=41*40+41=41^2 OR p=42*41+41=41*43.....NOT a prime

(2) p < 43
p could be anything

Combined
 p = n^2 - n + 41....n(n-1)+41<43.......n^2-n-2<0.....(n-2)(n+1)<0 
So n can be 0 or 1.
And answer will be YES always
Suff

C

GCMEMBER · Answer

Is the positive integer p prime?
 
(1) p = n^2 - n + 41, where n is an integer

p = n(n-1) + 41
For n = 1, p = 41, which is prime.

For n = 41, 
p = 41(41-1) + 41
= 41*40 + 41 
=41*41
which is not prime

Not sufficient.

(2) p < 43
p = 2, which is prime
p = 4, which is not prime

Not sufficient.

(1) + (2)
From (1), p=41, 43, 47 for different values of n.
From (2), p=42, 41, 40, 39 etc.

Only 41 is common between the two.
Therefore, p = 41

Hence, sufficient.

Option C

TestPrepUnlimited · Answer

Statement 1: 
The variable part  n^2 - n = n*(n - 1)  can be 1*0 = 0, 2*1 = 2, 2*3 = 6, etc. If we plug in negative n's we get those numbers too as Negative*Negative=Positive. 
Then p can be =41, 43, 47 etc. It looks like we have all primes but we can plug in a multiple of 41 for n so that p is a multiple of 41, and p would not be prime. Insufficient.

Statement 2: 
Insufficient.

Combined: 
The list earlier has only p = 41 less than 43, since there's only one possible p this is sufficient.

Ans: C

avigutman · Answer

Start with S2: positive integer p is between 1 and 42. Some, but not all, of those are prime. Eliminate BD.

S1 should be rephrased: p = n(n-1) + 41 where n is an integer and p is positive. This means that p is 41 units to the right of the product of 2 consecutive integers (which can't be negative, since n and (n-1) cannot be on opposite sides of zero for any integer value of n). n(n-1) could be 0, 2, 6, 12, 20, etc.
If n(n-1) = 0, then yes 0+41 is prime. If n(n-1) is a multiple of 41 (e.g. if n = 41 or n=42), then p is not prime. Eliminate A.

Combined, n(n-1) has to be 0, which means that p=41. Sufficient.

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Is the positive integer p prime? (1) p = n^2 - n + 41, where n is an

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