For every positive integer x, f(x) represents the greatest prime factor of x!, and g(x) represents the smallest prime factor of (2^x)+1. What is (g(f(12))?

Here, the output of f(x) will give the largest prime number less than X or X itself if X is a prime number. (Question uses complex language to confuse people.)

The output of g(x) will be the smallest prime factor. (which will be greater than 2 since the output of g(x) will be odd. )

Now,

f(12) = 11

and g(11) = 3 (2^11 + 1 = 2048 + 1 = 2049 , which is divisible by 3, The smallest possible prime factor.)


ANSWER : B

For every positive integer \(x\), \(f(x)\) represents the greatest prime factor of \(x!\), and \(g(x)\) represents the smallest prime factor of \(2^x+1\). What is \((g(f(12))\)?

f(x) = greatest prime factor of (x!)
f(12) = greatest prime factor of (12!) --> greatest prime factor of (12x11x10x...x2x1) = 11

g(x) = smallest prime factor of (2^x+1)
g(f(12)) = g(11) = smallest prime factor of (2^11+1) = smallest prime factor of (2049) = smallest prime factor of (3*683) = 3

Therefore g(f(12)) = 3.

A. 2
B. 3
C. 5
D. 7
E. 11

Answer Choice: B

So f(x) = greatest prime factor of x!
g(x)= smallest prime factor of (2^x+1)
Now asked g(f(12)) =?

f(12) = 1•2•3•4•5•6•7•8•9•10•11•12
Greatest prime factor =11
.: f(12)=11

g(11) = 2^11 + 1 => 2^5•2^6+1
g(11) = 2049 = 3•683
Smallest prime factor =3
.: g(f(12) =3

Answer B

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The largest prime factor for \(12!\) is \(11\), so\(f(12) = 11\)
if \(x = 11\), then\(2^{11} + 1 = 2048 + 1 = 2049\), not divisible by 2, but divisible by 3, so \((g(f(12)) = 3\)

B
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f(x) represents the greatest prime factor of x!.
12!=1*2*3*4*5......9*10*11*12
Distinct Prime factors of 12! are 2,3,5,7 and 11. Hence the greatest prime factor of 12! is 11

f(12)=11


g(x) represents the smallest prime factor of \(2^x+1\)
g(11) is the smallest prime factor of \(2^{11}+1\)

\(2^{11}\)+1 = 1 (Mod 2). Hence \(2^x+1\) is not divisible by 2.

\(2^{11}\)= \((-1)^{11}\) Mod 3
\(2^{11}+1\)= [(-1)+1] Mod 3
\(2^{11}\)+1= 0 Mod 3
\(2^{11}+1\) is divisible by 3. Hence the smallest prime factor of \(2^{11}+1\) is 3.

(g(f(12))=3

IMO B

For every positive integer xx, f(x)f(x) represents the greatest prime factor of x!x!, and g(x)g(x) represents the smallest prime factor of 2x+12x+1. What is(g(f(12))?

solve for (g(f(12))
f(12)= greatest prime factor of 12! ; 11
so g(11) = 2^(x+1) ; 2^11+1 ; 2048+1 ; 2049
so smallest prime factor ; 3

IMO B
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(g(f(12)))
f(12)= greatest prime factor of 12! This means the highest prime number from 1 to 12 which is 7.
Hence f(12)=7
Now g(7) is the least prime factor of ((2^7)+1)
(2^7)+1 = 128+1= 129.
Prime factors of 129 are
{3 and13}.
Hence g(7)=3.
The answer is therefore B.

Posted from my mobile device

For every positive integer x, f(x)represents the greatest prime factor of x!, and g(x) represents the smallest prime factor of 2^x+1. What is (g(f(12))?

A. 2
B. 3
C. 5
D. 7
E. 11

Given (g(f(12))
f(12) = highest prime factor of 12! will be 11 since there will no prime factors greater than 11
now g(11) = smallest prime factor of\(2^{11}\) +1

\(2^{11}\) will be even and +1 will make it odd
now \(2^{11}\) +1 = 2048 +1 =2049
this is completely divisble by 3 giving 683
thus smallest is 3 hence B

f(x) represents the greatest prime factor of x!
g(x) represents the smallest prime factor of \(2^x\)+1

(g(f(12))?

f(12) = 11 (greatest prime factor of 12!)

(g(11)) = 3 (smallest prime factor of 2^11+1 = 2049 = 3*683)


(g(f(12)) = 3

B is the answer

f(12) = greatest prime factor of 12!
The greatest prime factor of 12! is 11
f(12) = 11
g(11) = smallest prime factor of 2^11 + 1
= smallest prime factor of 2049 = 3


The answer is (B).
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