<< problem 129 - Repunit divisibility Prime cube partnership - problem 131 >>

# Problem 130: Composites with prime repunit property

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).

# My Algorithm

Copying getMinK from problem 129 works fine and shows the correct result after a few milliseconds.

However, due to the modified Hackerrank problem I had to change my algorithm from the ground up:
large numbers require a primality test that can handle number approx 10^12 → Miller-Rabin test from my toolbox

And it took me some time to realize that there is no need to find the smallest k:
if p-1 is a multiple of A(p) then R(A(p)) can be divided by p-1.
When I look at R(2 * A(p)) then it obviously has twice as many digits as R(A(p)).
But R(2 * A(p)) is a multiple of R(A(p)): R(2 * A(p)) = (10^p + 1) * R(A(p)).
Let's assume that A(p) = 2R(2) = 11 and R(2 * 2) = R(4) = 1111 = (10^2 + 1) * 11 = 101 * 11

Therefore I simply check whether R(p-1) (instead of R(A(p))) is a multiple of p.
And there is a pretty fast test for this:
R(p-1) = 111...111 = dfrac{10^{p-1}}{9} - 1
1 = dfrac{10^{p-1}}{9 * R(p - 1)}

Using powmod, which I need anyway for the Miller-Rabin test, I can find the residue of powmod(10, p-1, 9*p) extremely fast
→ if it's 1, then I have a match.

## Modifications by HackerRank

Aside from the larger input range, I have to print all matches instead of the sum of the first 25 matches.
But my code is still not good enough to solve all their test cases - I only get a 48% score.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Input data (separated by spaces or newlines):
Note: Enter the range of values to be analyzed.

This is equivalent to
echo "2 100" | ./130

Output:

(this interactive test is still under development, computations will be aborted after one second)

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

The code contains #ifdefs to switch between the original problem and the Hackerrank version.
Enable #ifdef ORIGINAL to produce the result for the original problem (default setting for most problems).

       #include <iostream>
#include <vector>

#define ORIGINAL

// ---------- Miller-Rabin test from my toolbox ----------

// return (a*b) % modulo
unsigned long long mulmod(unsigned long long a, unsigned long long b, unsigned long long modulo)
{
#ifdef __GNUC__
// use GCC's optimized 128 bit code
return ((unsigned __int128)a * b) % modulo;
#endif

// (a * b) % modulo = (a % modulo) * (b % modulo) % modulo
a %= modulo;
b %= modulo;

// fast path
if (a <= 0xFFFFFFF && b <= 0xFFFFFFF)
return (a * b) % modulo;

// we might encounter overflows (slow path)
// the number of loops depends on b, therefore try to minimize b
if (b > a)
std::swap(a, b);

// bitwise multiplication
unsigned long long result = 0;
while (a > 0 && b > 0)
{
// b is odd ? a*b = a + a*(b-1)
if (b & 1)
{
result += a;
if (result >= modulo)
result -= modulo;
// skip b-- because the bit-shift at the end will remove the lowest bit anyway
}

// b is even ? a*b = (2*a)*(b/2)
a <<= 1;
if (a >= modulo)
a -= modulo;

// next bit
b >>= 1;
}

return result;
}

// return (base^exponent) % modulo
unsigned long long powmod(unsigned long long base, unsigned long long exponent, unsigned long long modulo)
{
unsigned long long result = 1;
while (exponent > 0)
{
// fast exponentation:
// odd exponent ? a^b = a*a^(b-1)
if (exponent & 1)
result = mulmod(result, base, modulo);

// even exponent ? a^b = (a*a)^(b/2)
base = mulmod(base, base, modulo);
exponent >>= 1;
}
return result;
}

// Miller-Rabin-test
bool isPrime(unsigned long long p)
{
// IMPORTANT: requires mulmod(a, b, modulo) and powmod(base, exponent, modulo)

// some code from             https://ronzii.wordpress.com/2012/03/04/miller-rabin-primality-test/
// with optimizations from    http://ceur-ws.org/Vol-1326/020-Forisek.pdf
// good bases can be found at http://miller-rabin.appspot.com/

// trivial cases
const unsigned int bitmaskPrimes2to31 = (1 <<  2) | (1 <<  3) | (1 <<  5) | (1 <<  7) |
(1 << 11) | (1 << 13) | (1 << 17) | (1 << 19) |
(1 << 23) | (1 << 29); // = 0x208A28Ac
if (p < 31)
return (bitmaskPrimes2to31 & (1 << p)) != 0;

if (p %  2 == 0 || p %  3 == 0 || p %  5 == 0 || p % 7 == 0 || // divisible by a small prime
p % 11 == 0 || p % 13 == 0 || p % 17 == 0)
return false;

if (p < 17*19) // we filtered all composite numbers < 17*19, all others below 17*19 must be prime
return true;

// test p against those numbers ("witnesses")
// good bases can be found at http://miller-rabin.appspot.com/
const unsigned int STOP = 0;
const unsigned int TestAgainst1[] = { 377687, STOP };
const unsigned int TestAgainst2[] = { 31, 73, STOP };
const unsigned int TestAgainst3[] = { 2, 7, 61, STOP };
// first three sequences are good up to 2^32
const unsigned int TestAgainst4[] = { 2, 13, 23, 1662803, STOP };
const unsigned int TestAgainst7[] = { 2, 325, 9375, 28178, 450775, 9780504, 1795265022, STOP };

// good up to 2^64
const unsigned int* testAgainst = TestAgainst7;
// use less tests if feasible
if (p < 5329)
testAgainst = TestAgainst1;
else if (p < 9080191)
testAgainst = TestAgainst2;
else if (p < 4759123141ULL)
testAgainst = TestAgainst3;
else if (p < 1122004669633ULL)
testAgainst = TestAgainst4;

// find p - 1 = d * 2^j
auto d = p - 1;
d >>= 1;
unsigned int shift = 0;
while ((d & 1) == 0)
{
shift++;
d >>= 1;
}

// test p against all bases
do
{
auto x = powmod(*testAgainst++, d, p);
// is test^d % p == 1 or -1 ?
if (x == 1 || x == p - 1)
continue;

// now either prime or a strong pseudo-prime
// check test^(d*2^r) for 0 <= r < shift
bool maybePrime = false;
for (unsigned int r = 0; r < shift; r++)
{
// x = x^2 % p
// (initial x was test^d)
x = mulmod(x, x, p);
// x % p == 1 => not prime
if (x == 1)
return false;

// x % p == -1 => prime or an even stronger pseudo-prime
if (x == p - 1)
{
// next iteration
maybePrime = true;
break;
}
}

// not prime
if (!maybePrime)
return false;
} while (*testAgainst != STOP);

// prime
return true;
}

// ---------- getMinK from problem 129 ----------

// return minimum k where R(k) is divisible by x
unsigned long long getMinK(unsigned long long x)
{
// same as gcd(x, 10) = 1
if (x % 2 == 0 || x % 5 == 0)
return 0;

// "number of ones"
unsigned long long result  = 1;
// current repunit mod divisor
unsigned long long repunit = 1;
// no remainder ? that repunit can be divided by divisor
while (repunit != 0)
{
// next repunit
repunit *= 10;
repunit++;
// keep it mod divisor
repunit %= x;

result++;
}

return result;
}

int main()
{
unsigned long long from = 2;
unsigned long long to   = 15000;

#ifdef ORIGINAL
unsigned int maxFound = 25;
unsigned int numFound = 0;
unsigned int sum = 0;

#else
std::cin >> from >> to;
#endif

// for all even numbers gcd(i, 10) != 1 (it's >= 2)
if (from % 2 == 0)
from++;
// 91 is the first match
if (from < 91)
from = 91;

for (unsigned int p = from; p <= to; p += 2)
{
// reject prime numbers
if (isPrime(p))
continue;

// my old code based on problem 129:
// find minimum k (can be zero if gcd(p, 10) == 1)
//auto k = getMinK(p);
//if (k == 0)
//continue;
//if ((p - 1) % k == 0)

// ==> replaced by much faster idea:
// don't look for the smallest k
// just test if R(i-1) is a multiple of i
if (powmod(10, p - 1, 9 * p) == 1)
{
#ifdef ORIGINAL
sum += p;
numFound++;
if (numFound == maxFound)
break;
#else
std::cout << p << std::endl;
#endif
}
}

#ifdef ORIGINAL
std::cout << sum << std::endl;
#endif

return 0;
}


This solution contains 39 empty lines, 59 comments and 13 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

July 17, 2017 submitted solution

# Hackerrank

My code solves 14 out of 26 test cases (score: 52%)

I failed 0 test cases due to wrong answers and 12 because of timeouts

# Difficulty

Project Euler ranks this problem at 45% (out of 100%).

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

# Similar problems at Project Euler

Problem 129: Repunit divisibility

Note: I'm not even close to solving all problems at Project Euler. Chances are that similar problems do exist and I just haven't looked at them.

# Heatmap

Please click on a problem's number to open my solution to that problem:

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

 << problem 129 - Repunit divisibility Prime cube partnership - problem 131 >>
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