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Looking at trivikram's solution, it appears that having p2 = 2 is important here...

Letting p1, p2 be the two primes,

s=p1+p2
p=p1*p2

p-s=(p1*p2)-(p1+p2)
p-s=(p1*p2)-p1-p2

Since you have to subtract (p1+p2) from their product, it is important that you don't make the product too big, otherwise p-s would be too far from p1 chosen. The smallest prime number is 2, so p2=2.

In picking a number for p1, you want to pick a number that when you add 2, is the next prime number of a number in the answer choice. So that when you subtract (p1+p2) from (p1*p2) you get the number in the answer choice (p1-2).

Ummm....

Basically, you want to see if by adding 2 to a number in the answer choice, you get a prime number.

The only choice that does not meet this requirement is 119, for 119+2=121 which is divisible by 11.

What I wrote above appears pretty much an ANALYSIS of the answer rather than a mathematically justified approach to solve this problem...

I have this feeling that I'm circumventing...I know I'm heading in the right direction(?) but not quite getting there...

now make one of the prime number constant and othe variable, lets say Y=2 (prime number). we have x(2-1)-2 = X-2 now put different values of prime numbers in place of X to arrive at the solution.

now make one of the prime number constant and othe variable, lets say Y=2 (prime number). we have x(2-1)-2 = X-2 now put different values of prime numbers in place of X to arrive at the solution.

regards,

Amardeep

wow! thats cool. -)

i am wondering how to solve it within 2 minuts during test...

only analysis of the x-2 = 37, 121, 163, 353, 601 numbers could take much time ...

of course i know that 121 is not prime, but i would have to check others...

following are the basic tricks to be quicker in maths calculations:

1. take 5 pair of 3 digit numbers and multiply them. calculate your time taken per question (u need stop watch)... ideally u shouldnt take more than 21 secs per multipications ( e.g. 367*765)

following are the basic tricks to be quicker in maths calculations:

1. take 5 pair of 3 digit numbers and multiply them. calculate your time taken per question (u need stop watch)... ideally u shouldnt take more than 21 secs per multipications ( e.g. 367*765)

Re: If s is the sum of two prime numbers and p is the product of [#permalink]
27 May 2014, 05:23

2

This post was BOOKMARKED

kevincan wrote:

If S is the sum of two prime numbers and P is the product of these prime numbers, which of the following could NOT be the value of P - S ?

(A) 35 (B) 119 (C) 161 (D) 351 (E) 397

If S is the sum of two prime numbers and P is the product of these prime numbers, which of the following could NOT be the value of P - S ?

(A) 35 (B) 119 (C) 161 (D) 351 (E) 397

Let the prime numbers be \(a\) and \(b\). Given: \(P=a*b\) Given: \(S=a+b\)

\(P - S = ab - (a+b) = ab - a - b\)

The key here is to rewrite this as: \(P - S = ab - a - b = ab - a - b +1 -1 = (a - 1)*(b - 1) - 1\)

Now, we need to find the answer choice that CANNOT be written in this form, so if it can be, it's wrong.

Perhaps the best way to go about this is to add one to all the answer choices and then see if it can be written in the form of \((a - 1)*(b - 1)\), remembering that both \(a\) and \(b\) are prime. Essentially, we are going to write the ways we can express (answer choice +1) as a product of two factors, and if we can find a combination of two factors that are both one less than a prime number, it is the wrong choice.

A. \(35+1=36= 6*6=(7-1)*(7 - 1)\)--> 7 is prime, so this choice is incorrect.

B. \(119+1=120=12*10=(13-1)*(11-1)\) --> 13 and 11 are both prime, so incorrect

C. \(161+1=162=1*162= (2-1)*(163-1)\) --> 2 and 163 are both prime, so incorrect

D. \(351+1 = 352 = 32*11 = 16*22 =(17-1)*(23-1)\) --> 17 and 23 are both prime, so incorrect.

E. \(397+1 = 398 = 1*398 = 2*199\) --> This must be the correct answer because we eliminated everything else, but also because \(2*399\) and \(3*200\) are both in the form \(prime*composite\) and there are no other factors combinations.

Answer E.

gmatclubot

Re: If s is the sum of two prime numbers and p is the product of
[#permalink]
27 May 2014, 05:23

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