For how many positive integers n is n^2 - 3n + 2 a prime number?

Question

(A) none
(B) one
(C) two
(D) more than two, but finitely many
(E) infinitely many

EgmatQuantExpert · Accepted Answer

Solution 
  Given:  
 • The number n is a positive integer
• The expression  n^2 – 3n + 2  is a prime number

To find:  
 • The number of values possible for n

Approach and Working:  
If n = even, 
 • then  n^2 – 3n + 2  = even – even + even = even

if n = odd,
 • then  n^2 – 3n + 2  = odd – odd + even = even

Hence, for any value of n, the expression results in an even prime number.

We also know that 2 is the only even prime number.
 • Therefore,  n^2 – 3n + 2 = 2 
Or,  n^2 – 3n = 0 
Or,  n = 3   (as n cannot be 0)

Hence, the correct answer is option B.

Answer: B

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-NP-e1533292144892.png

Archit3110 · Answer

we can reqrite given expression as 
n^2 - 3n + 2
n(n-2)-1(n-2)
n=2,1
prime no 2 only 1
IMO B

stne · Answer

But where does it say that the given exp. is equal to zero ? The exp. can at max.be reduced to (n-2)(n-1), this does not necessarily mean that n =2 or n=1 , the variable n can take an infinite number of values. Unless and until we know the RHS of the exp. we cannot deduce the value of n. Hope you agree.

The method given above by EgmatQuantExpert uses a better logic,I suppose.

The method given above by  EgmatQuantExpert  uses a better logic,I suppose.

Again let me know if I have missed anything. Thanks.

IanStewart · Answer

Yes, you're right that the expression is not equal to zero. But the question asks if the number is prime, and so we care whether we can factor the number. If our number is equal to (n-2)(n-1), and if those factors are positive, our number could only be prime if one of those factors is equal to 1 -- otherwise we would have written our number as a product of two smaller numbers, which means our number is not prime. And if one of those factors is equal to 1, it must be the smaller of the two factors, so n -2 = 1, and n = 3. The product would also be prime if n = 0, but we're told n is positive here.

It is true that you can reach the answer to this question by noticing that the expression is always even, but that won't work in general. When you're asked if a number is prime, you really care if you can factor the number, not whether it's even or odd. If the question had asked instead about n^2 + 4n + 3, say, then an even/odd analysis wouldn't get you very close to an answer.

Crytiocanalyst · Answer

Solving we get (n-2)*(n-1)

and the only prime number that we can deduce for n=3 
Therefore IMO B

Pragya9s · Answer

Here is how I attempted this Question & then got it with the answer above by  EgmatQuantExpert ,

Equation n^2 - 3n +2 , we need to find out for how many positive integers does the equation is a Prime Number?

n = 1,
1 -3 +2 = 0

n = 2, 
4 - 6 +2 = -2

n = 3, 
9-9+2 = 2

n = 4, 
16 - 12 +2 = 6

n = 5, 
25 - 15+2 = 12

& so on...

From this we can deduce that for every odd or even value of 'n' we will get a Even number for the given equation. 
( n= even , even - even + even = even no. , n = odd , odd - odd + even = even )

The only Even Prime Number is 2 & hence option B is correct.

pierjoejoe · Answer

can't we just solve the equation:
n^2 - 3n +2 --> (n-1)(n-2)
this means that is the product of 2 consecutive numbers. two consecutive numbers have no common factor. 
so we have to find a prime that is the product of 2 consecutive numbers. obviously, by definition, a prime can have only 1 and itself as a factor, so we can exclude any prime number but 2. Because 2 is the only prime that is the product of 2 consecutive numbers.

bumpbot · Answer

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For how many positive integers n is n^2 - 3n + 2 a prime number?

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For how many positive integers n is n^2 - 3n + 2 a prime number?

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