<< problem 587 - Concave triangle Marsh Crossing - problem 607 >>

# Problem 601: Divisibility streaks

For every positive number n we define the function streak(n)=k as the smallest positive integer k such that n+k is not divisible by k+1.
E.g:
13 is divisible by 1
14 is divisible by 2
15 is divisible by 3
16 is divisible by 4
17 is NOT divisible by 5
So streak(13)=4.

Similarly:
120 is divisible by 1
121 is NOT divisible by 2
So streak(120)=1

Define P(s,N) to be the number of integers n, 1<n<N, for which streak(n)=s.
So P(3,14)=1 and P(6,10^6)=14286

Find the sum, as i ranges from 1 to 31, of P(i,4^i).

# My Algorithm

As always, I wrote a bruteForce function to solve the examples.

The equations for streak(13) can be written as:
13 == 0 mod 1
(13+1) == 0 mod 2
(13+2) == 0 mod 3
(13+3) == 0 mod 4
(13+4) != 0 mod 5

Or in general for each streak(n)=6:
n == 0 mod 1
(n+1) == 0 mod 2
(n+2) == 0 mod 3
(n+3) == 0 mod 4
(n+4) == 0 mod 5
(n+5) == 0 mod 6
(n+6) != 0 mod 7

The equations become a lot simpler if I replace n by its predecessor m = n - 1:
(m+1) == 0 mod 1
(m+2) == 0 mod 2
(m+3) == 0 mod 3
(m+4) == 0 mod 4
(m+5) == 0 mod 5
(m+6) == 0 mod 6
(m+7) != 0 mod 7

Since (m + x) mod x is the same as m mod x I can write:
m == 0 mod 1
m == 0 mod 2
m == 0 mod 3
m == 0 mod 4
m == 0 mod 5
m == 0 mod 6
m != 0 mod 7

A number is divisible by 1,2,3,4,5,6 if it is a multiple of the least common multiple lcm(1,2,3,4,5,6) = 60.
Each n with a predecessor that is such a multiple is a solution to this problem.

Thus I could compute 4^i / lcm(1,2,3,...,i) but there is a catch: it includes longer streaks, too.
Therefore I need 4^i / lcm(1,2,3,...,i,i+1) as well because that's counting the numbers with these longer streaks.
All divisions need to be integer divisions, discarding any fractional digits.

It took me a few minutes figuring out that 1 and 4^i are NOT part of the range of valid numbers (therefore minus 2).
→ that's why I keep on writing those bruteForce() function, they detect those anomalies easily.

For example:
P(6, 10^6) = \lfloor dfrac{10^6 - 2}{lcm(1,2,3,4,5,6)} \rfloor - \lfloor dfrac{10^6 - 2}{lcm(1,2,3,4,5,6,7)} \rfloor

P(6, 10^6) = \lfloor dfrac{999998}{60} \rfloor - \lfloor dfrac{999998}{420} \rfloor

P(6, 10^6) = 16666 - 2380
P(6, 10^6) = 14286

# 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 highest exponent i of 4^i

This is equivalent to
echo 10 | ./601

Output:

Note: the original problem's input 31 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.

       #include <iostream>

// ---------- code from my toolbox ----------

// greatest common divisor
template <typename T>
T gcd(T a, T b)
{
while (a != 0)
{
T c = a;
a = b % a;
b = c;
}
return b;
}

// least common multiple
template <typename T>
T lcm(T a, T b)
{
// parentheses avoid overflows for certain input values
return a * (b / gcd(a, b));
}

// ---------- problem-specific code ----------

// count numbers up to limit with a given streak
unsigned long long bruteForce(unsigned long long limit, unsigned int streak)
{
unsigned long long result = 0;
for (unsigned long long i = 2; i < limit; i++)
{
unsigned int current = 1;
while ((i + current - 1) % current == 0)
current++;

// counted one too far (the number after the streak ended)
current--;
// correct streak ?
if (current == streak)
result++;
}

return result;
}

// find result faster than the blink of an eye ...
unsigned long long solve(unsigned long long limit, unsigned int streak)
{
// find least common multiple of 1..streak
unsigned long long multiple = streak;
for (unsigned long long i = 2; i < streak; i++)
multiple = lcm(multiple, i);

// each n - 1 which is a multiple of "multiple" is valid solution
limit--;
// the last number is excluded from the range, too
limit--;
auto atLeast = limit / multiple;

// but there are a few number with a longer streaks !
multiple = lcm(multiple, streak + 1ULL);
auto tooMany = limit / multiple;

// subtract them
return atLeast - tooMany;
}

int main()
{
//std::cout << bruteForce( 14, 3) << std::endl;
//std::cout << bruteForce(100, 6) << std::endl;

unsigned int limit = 31;
std::cin >> limit;

unsigned long long result = 0;
unsigned long long pow4   = 4;
for (unsigned int i = 1; i <= limit; i++)
{
//result += bruteForce(last, i);
result += solve(pow4, i);
pow4 *= 4;
}

// another problem solved !
std::cout << result << std::endl;
return 0;
}


This solution contains 15 empty lines, 18 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

October 10, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 20% (out of 100%).

# 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 587 - Concave triangle Marsh Crossing - problem 607 >>
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