What's so special about 341?

Number Theory Level 4

For all positive, integral, non-pseudoprime a ≥ 2 a \geq 2 that satisfy

a ∣ 2 a − 2 a \mid 2^a-2

compute the sum of all factors of a a in terms of a a .

a 3 − 2 a a^3-2a No such a a exists a + 1 a+1 a 2 + 2 a + 12 a^2+2a+12

Finn Hulse by Finn Hulse , 21, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Nguyen Thanh Long
Jul 21, 2014

According to the little fermat theorem: a ∣ 2 × ( 2 a − 1 − 1 ) ⇔ a | 2 \times (2^{a-1} -1) \Leftrightarrow a=2 or a is a prime number. So a has two distinct factors: 1 and a. R e s u l t = a + 1 Result = \boxed{a+1}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Nathan Ramesh
May 27, 2014

Obviously a=2 and the only option that works is a+1.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Dude look deeper into this problem. As a matter of fact, all primes work along with a couple of pseudoprimes .

Finn Hulse - 7 years ago

Log in to reply

exaclty, finn.. all primes work. and factors of a prime are 1 and number itself. so, sum is a+1.

Pradeep Ch - 7 years ago

Dude, I solved all the problems except the third one of this set. People have reported it too. I would like if you could explain. @Finn Hulse

Krishna Ar - 7 years ago

The pseudo primes are basically the Carmichael Numbers, the outliers in Fermat's Prime Testing.

Nanayaranaraknas Vahdam - 7 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...