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Re: A function g(n), where n is an integer, is defined as the product of [#permalink]
Is option 2 the answer because when you add an even number to an even number it will always be even and hence cannot be prime.
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A function g(n), where n is an integer, is defined as the product of [#permalink]
Dillesh4096 wrote:
Bunuel wrote:
A function g(n), where n is an integer, is defined as the product of all integers from 1 to n. How many of the followings must be a prime number?

g(11) + 5; g(11) + 6; g(11) + 7; and g(11) + 8?


A. 1
B. 2
C. 3
D. 4
E. None


g(n) = 1*2*3.....*n = n!
--> g(11) = 11!

Note: Observe that g(11) = 11! can be written as 5A or 6B or 7C or 8D, where A,B,C & D are positive integers

g(11) + 5 = 5A + 5 = 5(A + 1) --> Not Prime
g(11) + 6 = 6B + 6 = 6(B + 1) --> Not Prime
g(11) + 7 = 7C + 7 = 7(C + 1) --> Not Prime
g(11) + 8 = 8D + 8 = 8(D + 1) --> Not Prime

IMO Option E

Pls Hit Kudos if you like the solution


Hello Dillesh4096

I know this are called co-primes and thats why they are not primes but, what happen if for example we have:

5! + 7 = 127 which is prime

How can we know if a number is prime or not, 5! is easy cuz we know thevalue but with bigger numbers?

Kind regards!
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A function g(n), where n is an integer, is defined as the product of [#permalink]
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