<< problem 68 - Magic 5-gon ring Totient permutation - problem 70 >>

# Problem 69: Totient maximum

Euler's Totient function, phi(n) [sometimes called the phi function], is used to determine the number of numbers less than n which are relatively prime to n.
For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, phi(9)=6.

nRelatively Primephi(n)n/phi(n)

2112
31,221.5
41,322
51,2,3,441.25
61,523
71,2,3,4,5,661.1666...
81,3,5,742
91,2,4,5,7,861.5
101,3,7,942.5

It can be seen that n=6 produces a maximum n/phi(n) for n <= 10.

Find the value of n <= 1000000 for which n/phi(n) is a maximum.

# Algorithm

I have no formal proof yet (it's too late - time to go to bed !):
in my first brute-force attempt I observed that the "best number" is the product of all primes
best = 2 * 3 * 5 * 7 * 11 * 13 * ... where best < 1000000.

Simple tests showed that all primes from 2 to 57 are sufficient.

## Modifications by HackerRank

The test best * nextPrime >= limit might overflow.
The same result can be achieved this way:
best * nextPrime >= limit

best >= dfrac{limit}{nextPrime}

All variables are integers and thus rounding comes into play.
The correct formula is:
best >= dfrac{limit}{nextPrime} + dfrac{nextPrime - 1}{nextPrime}

best >= dfrac{limit + nextPrime - 1}{nextPrime}

# My code

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

       #include <iostream>

int main()
{
// enough primes for this problem
const unsigned int primes[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 57 };

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

// multiply until we hit our limit
unsigned long long bestPos = 1;
for (auto p : primes)
{
// continue until bestPos reaches or exceeds our input value
//__int128 next = bestPos * p;
//if (next >= limit)
//  break;
// same code as before but more portable:
if (bestPos >= (limit + p - 1) / p)
break;

bestPos *= p;
}

std::cout << bestPos << std::endl;
}
return 0;
}


This solution contains 5 empty lines, 7 comments and 1 preprocessor command.

# 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 "1 10" | ./69

Output:

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

# Benchmark

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

March 1, 2017 submitted solution

# Hackerrank

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

# Difficulty

Project Euler ranks this problem at 10% (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 never an option.

projecteuler.net/thread=69 - the best forum on the subject (note: you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-69-find-the-value-of-n-≤-1000000-for-which-nφn-is-a-maximum/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p069.java (written by Nayuki)
Go: github.com/frrad/project-euler/blob/master/golang/Problem069.go (written by Frederick Robinson)
Scala: github.com/samskivert/euler-scala/blob/master/Euler069.scala (written by Michael Bayne)

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are already solved and there wasn't a Hackerrank version of it (at the time I solved it) or I didn't care about it because it differed too much.

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

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 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200
The 133 solved problems had an average difficulty of 16.9% at Project Euler and I scored 11,174 points (out of 12300) at Hackerrank's Project Euler+.
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