Repunit

26
Jan
2013
Project Euler 133: Repunit nonfactors

Project Euler 133: Repunit nonfactors

Problem 133 of Project Euler is a continuation of Problem 132 and Problem 129 in which we are supposed to find the some prime numbers which are not factors of R(10n) for any n. In fact the problem reads

A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.

Let us consider repunits of the form R(10n).

Although R(10), R(100), or R(1000) are not divisible by 17, R(10000) is divisible by 17. Yet there is no value of n for which R(10n) will divide by 19. In fact, it is remarkable that 11, 17, 41, and 73 are the only four primes below one-hundred that can be a factor of R(10n).

Find the sum of all the primes below one-hundred thousand that will never be a factor of R(10n).

Continue reading →

Posted by Kristian in Project Euler, 6 comments
19
Jan
2013
Project Euler 132: Large repunit factors

Project Euler 132: Large repunit factors

In problem 132 of Project Euler we are going back to working with repunits in a problem that reads

A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k.

For example, R(10) = 1111111111 = 11 x 41 x 271 x 9091, and the sum of these prime factors is 9414.

Find the sum of the first forty prime factors of R(109).

Continue reading →

Posted by Kristian in Project Euler, 4 comments
05
Jan
2013
Project Euler 130: Finding composite values, n, for which n−1 is divisible by the length of the smallest repunits that divide it.

Project Euler 130: Finding composite values, n, for which n−1 is divisible by the length of the smallest repunits that divide it.

Problem 130 of Project Euler is a continuation of problem 129, so if you have already solved that this one should be pretty easy. It reads

A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.

Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there always exists a value, k, for which R(k) is divisible by n, and let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = 5.

You are given that for all primes, p > 5, that p – 1 is divisible by A(p). For example, when p = 41, A(41) = 5, and 40 is divisible by 5.

However, there are rare composite values for which this is also true; the first five examples being 91, 259, 451, 481, and 703.

Find the sum of the first twenty-five composite values of n for which
GCD(n, 10) = 1 and n – 1 is divisible by A(n).

Continue reading →

Posted by Kristian in Project Euler, 2 comments
29
Dec
2012
Project Euler 129: Investigating minimal repunits that divide by n.

Project Euler 129: Investigating minimal repunits that divide by n.

Problem 129 of Project Euler reads

A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.

Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there always exists a value, k, for which R(k) is divisible byn, and let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = 5.

The least value of n for which A(n) first exceeds ten is 17.

Find the least value of n for which A(n) first exceeds one-million.

Continue reading →

Posted by Kristian in Project Euler, 7 comments