[Guest]

F(p) denotes the product all possible distinct primes less than p, whenever p > 3. Determine all possible value(s) of p that satisfy this equation:

F(p) = 2*p + 16

Edited October 5, 2009 by K Sengupta

[Guest]

p=7?

[Guest]

Agree with Ogden_tbsa

Let a = 7

and lets say there exists a prime b >7 which holds this identity

Then F(b) - F(a) = 2*(b-a)

F(b) = F(a)* 7.11... n such that n is <b

Let 7.11...n = X

then

F(a) (X-1) = 2*(b-a)

Then, 15(X - 1) = b-a

15X = b + 8

Since b-a does not increase as fast as LHS, there can be no other prime that satisfies this eqaution

Edited October 5, 2009 by DeeGee

[Guest]

Hi DG,

Didn't quite understand the following, can you explain?

F(b) = F(a)* 7.11... n such that n is <b

[Guest]

Answer is 7.

For p=7, all primes less than 7 are 2, 3 & 5.

so 2*3*5 = 30

and

2*7 + 16 = 30

[Guest]

Hi DG,

I really dont understand your logic of F(b) - F(a) = 2*(b-a)

say b is 89 and a is 7 (as you say)

so F(89)-F(7) = 2*82

so F(89) = 164-30 = 134

F(89) will be 2*3*5*7*11*13 ...... never be 134

[Guest]

It seems that I wasn't very clear in explaining the approach.

So here it is again with more explanation:

a = 7

so, F(a) = 2*3*5 = 30

For any prime "b" greater than 7, say 17, F(17) = 2*3*5*7*11*13 = F(a)*7*11*13

So, methinks, n in this case is 13; in case of 19 it would be 17 and so on

Then, F(17) - F(7) = F(7)*(7*11*13 - 1)

If F(17) holds for the equality F(p) = 2*p + 16 then,

F(7)*(7*11*13 - 1) must be equal to 2*(17 - 7) which is clearly not the case; in fact it will not be true for any prime greater than 7.

So, atuljain, the approach was to disprove the assumption that there exists a prime greater than 7 for which it would be true. Of course as you said, "F(89) will never be 134", the approach I mentioned is valid (I think) for all primes greater than 7

Hope it is clearer now!

[plainglazed]

It seems that I wasn't very clear in explaining the approach.

So here it is again with more explanation:

a = 7

so, F(a) = 2*3*5 = 30

For any prime "b" greater than 7, say 17, F(17) = 2*3*5*7*11*13 = F(a)*7*11*13

So, methinks, n in this case is 13; in case of 19 it would be 17 and so on

Then, F(17) - F(7) = F(7)*(7*11*13 - 1)

If F(17) holds for the equality F(p) = 2*p + 16 then,

F(7)*(7*11*13 - 1) must be equal to 2*(17 - 7) which is clearly not the case; in fact it will not be true for any prime greater than 7.

So, atuljain, the approach was to disprove the assumption that there exists a prime greater than 7 for which it would be true. Of course as you said, "F(89) will never be 134", the approach I mentioned is valid (I think) for all primes greater than 7

Hope it is clearer now!

As I read it, the OP does not say P must be a prime; just that F(P) is the product of all primes less than P. Have you proved that there is not some prime out there so much greater than the next lower prime that there is not some integer P that satisfies the equation? I think the key must be the 16. Still pondering...

[plainglazed]

As DeeGee notes combined with the proof of Bertrand's postulate the function F(p) will always be greater than 2p + 16 for all values of p greater than 7.

I think...

[Guest]

[snip]

For any prime "b" greater than 7, say 17, F(17) = 2*3*5*7*11*13 = F(a)*7*11*13

So, methinks, n in this case is 13; in case of 19 it would be 17 and so on

[snip]

Thanks! I thought (stupidly) that 7.11.. is a fractional number of the form 7.111111.. That threw me off completely.