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Number Theory Level 4

Find sum of all the primes of the form ${ n }^{ n }+1$ , where $n$ is an integer, which are less than ${ 10 }^{ 19 }$ .

If the answer is $x$ , submit your answer as $x + (x \bmod {33} )$ .

2 solutions

Sauditya Yo Yo
Jun 2, 2015

this number should be of form 2^(2^(2..........) +1 therefore answer is 2+5+257=264 (264=0(mod33))

Aditya Kumar
May 31, 2015

For $n=1 ,n=2$ , we get primes. An odd $n>1$ yields an even ${n^n+1}>2$ . So $n$ must be even, i.e., $n={ 2 }^{ { 2 }^{ t }(2k+1) }$ . Since

${ 2 }^{ { 2 }^{ t } }+1|{ 2 }^{ { 2 }^{ t }(2k+1) }+1$

The exponent of $n$ cannot have an odd divisor. Thus $n={ 2 }^{ { 2 }^{ t } }$ .

Therefore, we get $x=264$ .

Pi Han Goh - 6 years ago

but u asked to submit the answer as x+(xmod33). so it should be 561

D K - 6 years ago

Well, 264 is divisible by 33. So, 264mod33 =0. So the answer remains 264!!!

Didarul Azam - 6 years ago