tar013 wrote:Can someone explain to me the logic behind the answer?

f x is a prime number, the function G(x) is defined as the xth root of the product of all distinct primes less than or equal to x. If x is one of the first five primes, the maximum value of G(x) occurs when x =
(A) 2
(B) 3
(C) 5
(D) 7
(E) 11

Correct answer D

I think it would be a huge conflict of interest for me to post the answer to our own "Challenge Problem Showdown!!" But, let's start a dialogue -- any thoughts on how you'd set it up? What's your set of prime numbers you are considering?

Thanks!

-Brian

A. x = 2, G(2) = 2^(1/2)
B. x = 3, G(3) = (2*3)^(1/3) = 6^(1/3)
C. x = 5, G(5) = (2*3*5)^(1/5) = 30^(1/5)
D. x = 7, G(7) = (2*3*5*7)^(1/7) = 210^(1/7)
E. x = 11, G(11) = (2*3*5*7*11)^(1/11) = 2310^(1/11)

Note that G(2), G(3),... and G(11) are greater than 1.
Hence, whatever the order of them by value, that will not change if we raise them by same power.

Now, to make things easier we'll raise all of them by them LCM of the denominators of the powers, i.e. LCM of 2, 3, 5, 7, and 11 = 2310

[G(2)]^2310 = [2^(1/2)]^2310 = 2^1155
[G(3)]^2310 = [6^(1/3)]^2310 = 6^770 = (2^770)*(3^770) > (2^770)*(2^770) > 2^1540 > 2^1155

Hence, G(3) > G(2)

[G(5)]^2310 = [30^(1/5)]^2310 = 30^462
[G(7)]^2310 = [210^(1/7)]^2310 = 210^330 = (7^330)*(30^330) > ((√30)^330)*(30^330) > (30^165)*(30^330) > 30^495 > 30^462

Hence, G(7) > G(5)

[G(7)]^2310 = [210^(1/7)]^2310 = 210^330 = (6^330)*(35^330) > (6^330)*(6^330)*(5^330) > (6^330)*(6^330)*((√6)^330) > (6^330)*(6^330)*(6^165) > 6^825 > 66770

Hence, G(7) > G(3)

[G(11)]^2310 = [2310^(1/11)]^2310 = 2310^210 = (11*210)*(210^210) < ((√210)^210)*(210^210) < (210^105)*(210^210) < 210^315 < 210^330

Hence, G(11) < G(7)

Hence, the largest is G(7)

The correct answer is D.