Two prime numbers are considered consecutive if no other prime lies be

Question

Two prime numbers are considered consecutive if no other prime lies between them on the number line. If \(p_1\) and \(p_2\) are consecutive primes, with \(|p_1 – p_2| > 2\), what is the smallest possible absolute value of the coefficient of the x term in the distributed form of the expression \((x – p_1)(x – p_2)\)?

(A) 5
(B) 8
(C) 12
(D) 18
(E) 24

Kudos for a correct solution.

(A) 5
(B) 8
(C) 12
(D) 18
(E) 24

anudeep133 · Accepted Answer

Solution:   (x-p1)*(x-p2) = x^{2} - (p1 + p2) + p1p2 
The absolute co-efficient of x is p1+p2. And for this to be minimum and satisfy |p1–p2|>2, p1 and p2 should be 11 and 7.
So,p1+p2 = 18.

Option D

vp101 · Answer

I didn't really understand the question. Can someone explain it please?

I did however try to get to an answer. For example, the smallest two consecutive prime numbers that have a difference greater than two are 7 and 11. So I guessed that the answer is 12 since its the smallest number that is greater than both the primes - 7 and 11.

(12 - 7)(12 - 11) ??

Sorry if I'm not making any sense. Waiting for the OA...

Bambaruush · Answer

I am also puzzling with this problem. I find no connection between value of x and the end result of the given expression. Is it asking for minimum of both? If so there are 2 answers, 8 and 24, that qualifies with end result of -3.

(8-7)(8-11)=-3
(24-23)(24-27)=-3

(8-7)(8-11)=-3
(24-23)(24-27)=-3

sanderintelmann · Answer

We know that the set of prime numbers is 2, 3, 5, 7, 11, 13, 17, ...

Statement 1: p1 and p2 are consecutive primes, with |p1–p2|>2
--> the first set of two consecutive primes that qualifies is 7 and 11

Statement 2: smallest possible absolute value of the coefficient of the x term in the distributed form of the expression (x–p1)(x–p2)?
--> distributed form of (x–p1)(x–p2) --> (x-7)(x-11) --> x^2 - 7x - 11x +77 --> x^2 -18x + 77
--> smallest coefficient of the term x is therefore -18

anudeep133 · Answer

On what basis did you select 7 and 11 as qualified? The criteria of selecting p1 and p2 should be such that the value of p1+ p2 is minimum and |p1–p2|>2.Then only you get the values as 7 and 11. Can you explain me how you got 7 and 11 before getting into 2nd statement.?

sanderintelmann · Answer

You're right - you cannot evaluate statement 1 without reference to the constraint of "smallest possible absolute value." I went through the below pairs of primes when choosing 7 and 11. Given we are looking for the smallest possible coefficient of x, 7+11=18 is smaller than 13+17=30.

P1 P2 P1-P2 ABS >2?
2 3 -1 No
3 5 -2 No
5 7 -2 No
7 11 -4 Yes
11 13 -2 No
13 17 -4 Yes

anudeep133 · Answer

Its all fine then. It seemed to me that you selected 7,11 before statement 2. What you did is correct. Sry to disturb you

sunnysid · Answer

(x−p1)∗(x−p2)=x2−(p1+p2)x+p1p2

Value of (p1+p2) will be min. for the min consecutive primes which satisfy |p1-p2|>2

So the min. value p1 & p2 can take = 7, 11

p1+p2= 7+11= 18

Ans. D

harishbiyani8888 · Answer

Not sure of the answer. Waiting for Bunuel to post solution

GMATinsight · Answer

Let's solve the expression  (x – p_1)(x – p_2)  first

(x – p_1)(x – p_2)  =  x^2 - (p_1 + p_2)x + (p_1 * p_2)

Now, Co-efficient of x term    = (p_1 + p_2)

i.e. We need to calculate the Minimum Absolute value of  (p_1 + p_2)  which will be possible for MINIMUM values of both  p_1  &  p_2

But,  |p_1 – p_2| > 2  i.e. we need to find two consecutive SMALLEST Prime numbers which are separated by more than 2

List of Prime Numbers: 2, 3, 5,    7, 11   , 13, 17 ...

i.e.  p_1 = 7  and  p_2 = 11

i.e. Minimum Value of  (p_1 + p_2) = 7+11 = 18

Answer: option D

santorasantu · Answer

coefficient of x is (p1+p2). the sum of the smallest prime numbers that are consecutive and also with the condition  |p_1 – p_2| > 2  
are 7,11

first few primes are 2,3,5,7,11,13,17,19..
out of these 7 and 11 satisfy so 
D

kunal555 · Answer

Two prime numbers are considered consecutive if no other prime lies between them.
So the pairs of these prime numbers can be (2,3) (3,5) (5,7) (7,11) and so on
We are also given that difference between two prime numbers should be greater than 2.
Now, we are to find the smallest absolute value of coefficient of x in  (x – p_1)(x – p_2) 
Distributed for of this expression will be

(x – p_1)(x – p_2)  =  x^2 - p_1x - p_2x + p_1p_2  =  x^2 - (p_1+ p_2)x + p_1p_2

the coefficient of x here is  - (p_1+ p_2) 
We want smallest possible value of  |- (p_1+ p_2)|  or  (p_1+ p_2)

Our first such set of  p_1  and  p_2  according to the above mentioned condition will be (7,11)
So, smallest possible absolute value of the coefficient of the x term is 7+11 = 18

Answer:- D

Bunuel · Answer

MANHATTAN GMAT OFFICIAL SOLUTION:

The question seems forbidding, but start by grabbing onto the most concrete part, which comes at the end. Start with distributing  (x – p_1)(x – p_2) .
 (x – p_1)(x – p_2) = x^2 – (p_1 + p_2)x + p_1p_2

All we care about is the coefficient of the x term, which is  –(p_1 + p_2) . Specifically, we care about the absolute value of this, which is p_1 + p_2, since primes are by definition positive.

So what we are really asked for is the smallest possible value of  p_1 + p_2 , under two conditions:

1. These two primes are consecutive, meaning that there’s no other prime between them.
2.  |p_1 – p_2| > 2 , meaning that the primes are more than 2 units apart on the number line.

In other words, the question really is “what is the smallest possible sum of two consecutive primes that are more than 2 units apart?”

Now take the first several primes: 2, 3, 5, 7, 11, 13. The first pair of consecutive primes more than 2 units apart is {7, 11}. Their sum is 18.

The correct answer is D.

BrentGMATPrepNow · Answer

We're told that p1 and p2 are CONSECUTIVE primes (no primes in between)
We're also told that |p1 – p2| > 2, which means the two primes are MORE THAN 2 units away. 
So, for example, the two primes cannot be 3 and 5, since the two primes are 2 units away.

Can two CONSECUTIVE primes be THREE units away? 
For the primes to be THREE units away, one prime must be EVEN and the other must be ODD
Since 2 is the only EVEN prime, the smaller prime must be 2, which means the bigger prime must be 5. 
HOWEVER, 2 and 5 are not CONSECUTIVE primes, since the prime number 3 lies between 2 and 5
So, we can conclude that two consecutive primes be CANNOT be THREE units away

Can two CONSECUTIVE primes be FOUR units away?
Let's check some primes that are FOUR units away
3 and 7. No good. They are not CONSECUTIVE since the prime number 5 lies between 3 and 7.
5 and 9. No good. 9 isn't prime. 
7 and 11. PERFECT! 7 and 11 are prime, and they're consecutive.

So, 7 and 11 are the smallest values for p1 and p2.

What is the smallest possible absolute value of the coefficient of the x term in the distributed form of the expression (x – p1)(x – p2)?  
We can say that p1 = 7, and p2 = 11
We get: (x – p1)(x – p2) = (x – 7)(x – 11)
= x² - 11x - 7x + 77
= x²  - 18 x + 77
The COEFFICIENT of the x term is  -18

We want to find the smallest possible ABSOLUTE VALUE of the coefficient of the x term
So, we want | -18 |, which equals 18

Answer: D

Cheers,
Brent

SHYAM23 · Answer

p1=7 and p2 =11
(x-p1)(x-p2)
(x-7)(x-11)
x*x-7x-11x+11*7
x(x-7)-11(x-7)
x-11=0 or x-7=0
x=11
x=7
11+7=18

hadimadi · Answer

Hi,

I think this question is formulated unnecessarily complicated, maybe to confuse, but not to rest any crazy concepts. I might be mistaken as I am a non native English speaker, but whatever.

(x-p1)*(x-p1)=x^2-x(p1+p2)+p1*p2

The question asks for the smallest absolute value of the coefficient of the term with -x, which is the smallest absolute value of p1+p2. In other words, what is the smallest sum of two consecutive prime numbers which are farther than 2 away, or again in other words, what are the smallest two prime numbers which are farther than two away.

These are 7 and 11, and we get (D)

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Two prime numbers are considered consecutive if no other prime lies be

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