Number Theory
# Prime Numbers

Assume there are only a finite number of primes \(p_1,p_2,\ldots,p_n\). Let \(N\) be the product of all of those primes, add to it \(1\) and you get a

newprime number since it isn't divisible by any of the primes we listed at first. Contradiction! \(\Rightarrow\Leftarrow\) Therefore, there is an infinite number of primes.

Let \( q_1, q_2, q_3, \ldots \) be the list of all primes (in ascending order). Your mission is to find the smallest value of \( q_1 q_2 q_3 \ldots q_n + 1 \) that is not a prime.

Let \(A={102^1, 102^2, 102^3, \cdots}\). How many primes \(p\) are there such that \(A\) has at least one element \(a\) such that \(a \equiv -1 \text{ (mod p)}\)?

For example, one such prime is \(103\), because \(102^1 \equiv -1 \text{ (mod 103)}\).

\[\Large \frac{d}{dx} n \neq 0?\]

In calculus, when you take the derivative of a constant you get zero as an answer. In number theory, there is something called the arithmetic derivative which allows you to differentiate a number and get a nonzero answer. The arithmetic derivative works as follows.

Where \(n'\) denotes the arithmetic derivative of \(n\):

\(p' = 1\) for all primes \(p\)

\((ab)'=a'b+ab'\)

\(0'=1'=0\)

For example, \(6'=(2\times3)'=(2')(3)+(2)(3')=(1)(3)+(2)(1)=5\)

The double arithmetic derivative, denoted by \(n''\), is simply defined by \(n''=(n')'\).

Find the sum of all positive integers \(n<100\) such that \(n''=1\)

\[ \Huge { \sum_{m=1}^{2^{820}} } \large \left \lfloor \left \lfloor \frac{820}{1 + \displaystyle \sum_{j=2}^m \left \lfloor\frac{(j-1)!+1}{j} -\left \lfloor\frac{(j-1)!}{j}\right \rfloor \right \rfloor }\right \rfloor ^{1/820} \right \rfloor = \ ? \]

You may use this List of Primes as a reference.

Note: By definition, \( \displaystyle \sum_{a=b+1}^b 1 = 0 \).

What is the smallest prime number that cannot be written in the form \(\dfrac{pq + 1}{p + q}\), where \(p\) and \(q\) are prime numbers?

For example, \(2\) can be written as \(\dfrac{3*5 + 1}{3 + 5}\).

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