<< problem 11 - Largest product in a grid Large sum - problem 13 >>

# Problem 12: Highly divisible triangular number

The sequence of triangle numbers is generated by adding the natural numbers.
So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
The first ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

Let us list the factors of the first seven triangle numbers:

1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28

We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

# My Algorithm

Similar to other problems, my solution consists of two steps
1. precompute all possible inputs
2. for each test case: perform a simple lookup

It takes less than a second to find all such numbers with at most 1000 divisors.
Two "tricks" are responsible to achieve that speed:
You can get all divisors of x by analyzing all potential divisors i<=sqrt{x} instead of i<x.
Whenever we find a valid divisor i then another divisor j=frac{x}{y} exists.
The only exception is i=sqrt{x} because then j=i.

Somehow more subtle is my observation that when numbers have more than about 300 divisors,
the smallest one always end with a zero. I cannot prove that, I just saw it while debugging my code.

I decided to store all my results in a std::vector called smallest where
smallest[x] contains the smallest triangle number with at least x divisors.

While filling that container, the program encounters many "gaps":
e.g. 10 is the smallest number with 4 divisors and 28 is the smallest number with 6 divisors
but there is no number between 10 and 28 with 5 divisors.
Therefore 28 is the smallest number with at least 5 divisors, too.

## Alternative Approaches

Prime factorization can find the result probably a bit faster.

# Interactive test

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

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 7" | ./12

Output:

Note: the original problem's input 500 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++ and can be compiled with G++, Clang++, Visual C++. You can download it, too.

       #include <iostream>
#include <vector>

int main()
{
// find the smallest number with at least 1000 divisors
// (due to Hackerrank's input range)
const unsigned int MaxDivisors = 1000;

// store [divisors] => [smallest number]
std::vector<unsigned int> smallest;
smallest.push_back(0); // 0 => no divisors

// for index=1 we have triangle=1
// for index=2 we have triangle=3
// for index=3 we have triangle=6
// ...
// for index=7 we have triangle=28
// ...
unsigned int index    = 0;
unsigned int triangle = 0; // same as index*(index+1)/2
while (smallest.size() < MaxDivisors)
{
// next triangle number
index++;
triangle += index;

// performance tweak (5x faster):
// I observed that the "best" numbers with more than 300 divisors end with a zero
// that's something I cannot prove right now, I just "saw" that debugging my code
if (smallest.size() > 300 && triangle % 10 != 0)
continue;

// find all divisors i where i*j=triangle
// it's much faster to assume i < j, which means i*i < triangle
// whenever we find i then there is a j, too
unsigned int divisors = 0;
unsigned int i        = 1;
while (i*i < triangle)
{
// divisible ? yes, we found i and j, that's two divisors
if (triangle % i == 0)
divisors += 2;
i++;
}
// if i=j then i^2=triangle and we have another divisor
if (i*i == triangle)
divisors++;

// fill gaps:
// e.g. 10 is the smallest number with 4 divisors
//      28 is the smallest number with 6 divisors
// there is no number between 10 and 28 with 5 divisors
// therefore 28 is the smallest number with AT LEAST 5 divisors, too
while (smallest.size() <= divisors)
smallest.push_back(triangle);
}

unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned int minDivisors;
std::cin >> minDivisors;

// problem setting asks for "over" x divisors => "plus one"
std::cout << smallest[minDivisors + 1] << std::endl;
}

return 0;
}


This solution contains 9 empty lines, 24 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.5 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

February 23, 2017 submitted solution

# Hackerrank

My code solves 8 out of 8 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 5% (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.

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