<< problem 166 - Criss Cross Exploring the number of different ways a number ... - problem 169 >>

# Problem 168: Number Rotations

Consider the number 142857. We can right-rotate this number by moving the last digit (7) to the front of it, giving us 714285.
It can be verified that 714285 = 5 * 142857.
This demonstrates an unusual property of 142857: it is a divisor of its right-rotation.

Find the last 5 digits of the sum of all integers n, 10 < n < 10^100, that have this property.

# My Algorithm

Let's analyze the example: 714285 = 5 * 142857
If I replace each digit by a letter then: fabcde = x * abcdef where a=1, b=4, c=2, d=8, e=5, f=7 and x=5.

There is a special relationship between those variables:
e = (x * f) mod 10 = (5 * 7) mod 10 = 35 mod 10 = 5

The first digit of 35 - which is 3 - influences the value of d:
d = (x * e + 3) mod 10 = (5 * 5 + 3) mod 10 = 28 mod 10 = 8

The first digit of 28 - which is 2 - influences the value of c:
c = (x * d + 2) mod 10 = (5 * 8 + 2) mod 10 = 42 mod 10 = 2

... and so on.
Just by knowing x and f I can create all other variables.
However, not all produce a valid result: the product's first digit must be f, too.
Leading zeros are disallowed as well.

My program iterates for each number of digits numDigits over all possible values of x (parameter multiplier) and f (parameter lastDigit).
If the generated sequence doesn't produce a first digit that matches f then it returns 0.

The problem statement asks for the last five digits of all matching numbers, therefore a little modulo action is involved.

## Note

To me it wasn't clear whether multiplier can be 1. Because of this, many numbers with repeated digits are part of the solution
(e.g. 1111, 2222, 3333, ...).

# 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):

This is equivalent to
echo 10 | ./168

Output:

Note: the original problem's input 100 cannot be entered
because just copying results is a soft skill reserved for idiots.

(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. Or just jump to my GitHub repository.

       #include <iostream>

// return the unique n-digit number with a predefined last digit
// which after multiplication with a multiplier is a right-rotation of itself
// return only that value modulo some power-of-10
// if there is no such number, then return 0
unsigned int search(unsigned int numDigits, unsigned int multiplier, unsigned int lastDigit, unsigned int modulo)
{
// will be 10, 100, 1000, 10000, ...
unsigned int shift = 10;
// first digit of last iteration
unsigned int carry =  0;

// set last digit
unsigned int current = lastDigit;
unsigned int result  = lastDigit;

while (--numDigits)
{
// process next digit
auto next = multiplier * current + carry;
// "next" has at most two digits, carry over the first digit to next iteration
carry     = next / 10;
// and its second digit becomes part of "result"
current   = next % 10;

// current digit relevant for result ?
if (shift < modulo)
{
result += current * shift;
shift  *= 10;
}
}

// left-most digit
auto first = multiplier * current + carry;
// failed ? (no leading zero, product's first digit must match last digit of factor)
if (current == 0 || first != lastDigit)
return 0;

return result;
}

int main()
{
unsigned int maxDigits = 100;
std::cin >> maxDigits;

// last five digits
const unsigned int Modulo = 100000;

unsigned int result = 0;
// from 2 to 100 digits
for (unsigned int numDigits = 2; numDigits <= maxDigits; numDigits++)
// each multiplier between 1 and 9
for (unsigned int multiplier = 1; multiplier <= 9; multiplier++)
// last digit can't be zero, therefore 1 to 9 only
for (unsigned int lastDigit = 1; lastDigit <= 9; lastDigit++)
result += search(numDigits, multiplier, lastDigit, Modulo);

std::cout << result % Modulo << std::endl;
return 0;
}


This solution contains 10 empty lines, 17 comments and 1 preprocessor command.

# 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

August 3, 2017 submitted solution

# Hackerrank

My code solves 10 out of 10 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 65% (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.

# 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 166 - Criss Cross Exploring the number of different ways a number ... - problem 169 >>
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