If s is the sum of two prime numbers and p is the product of

lets assume X and Y are two prime numbers

P = X*Y and S = X+Y, so P-S = XY-X-Y or X(Y-1)-Y

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)

2. memorize squares upto 30 ( e.g. 1, 4, 9 , 16 etc..) cubes upto 15
(1,8,27.. etc).

3. prime numbers from 1 to 200.

Besides this you may go throuh vedic mathmatics to understand the tricky short-cuts ... its worth trying.

regards.

Amardeep

hey Vikram, what about 11 and 13. Product is 143, sum is 24, and difference is 119. B can be the difference of p-s. No?

Yep you are correct..I didnt see that combo.

So, if p1 and p2 are prime numbers, what do we know about their product minus their sum?

Amardeep, thanx a lot!

appreciate it.

You are right. These simple math operations require much time in the extreme environment.

whats to say the prime numbers can't be the same? i.e p1 = p2 = 2 ?

has anyone solved this problem

Kevincan, please give the solution to this problem.

I was almost stuck here but thanks to Amardeep who gave really a good solution

wait a minute!!!!

the value written for (E) option keeps changing in this post....

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

In trivikram's quote (A) 35 (B) 119 (C) 161 (D) 351 (E) 393

then again in trivikram's quote (A) 35 (B) 119 (C) 161 (D) 351 (E) 599

If we take the original poster's values then answer is B and E both

It is only if (E) 599, then we get answer B.

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.

Hi

Was anyone able to solve this sum within 2 mins?
Any other shorter way to do it?

Bunuel ,Karishma pls help!

Here x=Value *Y/Y-1
clearly only B doesnt satisfy Y being 2 is satisfied by all
Hence B

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