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

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If s is the sum of two prime numbers and p is the product of [#permalink]

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12 Mar 2007, 16:15
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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

Last edited by kevincan on 13 Mar 2007, 07:33, edited 1 time in total.
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12 Mar 2007, 16:48
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) 393

Should be B

Let the 2 primes be p1,p2

A) p1=37 , p2 =2 so s= 39 p = 74 p-s = 35

B) p1=121 , p2 =2 so s= 123 p = 242 p-s = 119 but

p1 =121 isnt prime

C) p1=163 and p2=2

D) p1=353 p2=2

E) p1=601 and p2=2

So B

My method is rustic but if anyone has a better way to do it please let me know
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12 Mar 2007, 17:51
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...

Help!
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13 Mar 2007, 00:01
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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
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13 Mar 2007, 00:53
Amardeep Sharma wrote:
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...
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13 Mar 2007, 01:40
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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
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13 Mar 2007, 04:04
trivikram wrote:
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) 599

Should be B

Let the 2 primes be p1,p2

A) p1=37 , p2 =2 so s= 39 p = 74 p-s = 35

B) p1=121 , p2 =2 so s= 123 p = 242 p-s = 119 but

p1 =121 isnt prime

C) p1=163 and p2=2

D) p1=353 p2=2

E) p1=601 and p2=2

So B

My method is rustic but if anyone has a better way to do it please let me know

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?
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13 Mar 2007, 04:34
aurobindo wrote:
trivikram wrote:
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) 599

Should be B

Let the 2 primes be p1,p2

A) p1=37 , p2 =2 so s= 39 p = 74 p-s = 35

B) p1=121 , p2 =2 so s= 123 p = 242 p-s = 119 but

p1 =121 isnt prime

C) p1=163 and p2=2

D) p1=353 p2=2

E) p1=601 and p2=2

So B

My method is rustic but if anyone has a better way to do it please let me know

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.
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13 Mar 2007, 07:34
So, if p1 and p2 are prime numbers, what do we know about their product minus their sum?
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13 Mar 2007, 19:32
Amardeep Sharma wrote:
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

Amardeep, thanx a lot!

appreciate it.

You are right. These simple math operations require much time in the extreme environment.
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14 Mar 2007, 21:39
whats to say the prime numbers can't be the same? i.e p1 = p2 = 2 ?

has anyone solved this problem
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06 Apr 2007, 16:43
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

Kevincan, please give the solution to this problem.
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06 Apr 2007, 17:01
I was almost stuck here but thanks to Amardeep who gave really a good solution
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06 Apr 2007, 17:13
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.
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Re: If s is the sum of two prime numbers and p is the product of [#permalink]

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27 May 2014, 06:23
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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.

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Re: If s is the sum of two prime numbers and p is the product of [#permalink]

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26 Sep 2015, 07:28
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Re: If s is the sum of two prime numbers and p is the product of [#permalink]

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13 Dec 2015, 05:36
Hi

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

Bunuel ,Karishma pls help!
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Re: If s is the sum of two prime numbers and p is the product of [#permalink]

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14 Mar 2016, 01:27
Here x=Value *Y/Y-1
clearly only B doesnt satisfy Y being 2 is satisfied by all
Hence B
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Re: If s is the sum of two prime numbers and p is the product of   [#permalink] 14 Mar 2016, 01:27
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