<< problem 119 - Digit power sum Disc game prize fund - problem 121 >>

# Problem 120: Square remainders

Let r be the remainder when (a-1)^n + (a+1)^n is divided by a^2.

For example, if a = 7 and n = 3, then r = 42: 6^3 + 8^3 = 728 == 42 mod 49.
And as n varies, so too will r, but for a = 7 it turns out that r_{max} = 42.

For 3 <= a <= 1000, find sum{r_{max}}.

# My Algorithm

The binomial expansion (see en.wikipedia.org/wiki/Binomial_theorem) is:

(x+y)^n = {{n}choose{0}} x^{n}y^0 + {{n}choose{1}} x^{n-1}y^1 + ... {{n}choose{n-1}} x^{1}y^{n-1} + {{n}choose{n}} x^{0}y^n

If x=a and y = \pm 1:
(a+1)^n = {{n}choose{0}} a^{n} + {{n}choose{1}} a^{n-1} + {{n}choose{2}} a^{n-2} + ... {{n}choose{n-1}} a^{1} + {{n}choose{n}} a^{0}
(a-1)^n = {{n}choose{0}} a^{n} - {{n}choose{1}} a^{n-1} + {{n}choose{2}} a^{n-2} - ... {{n}choose{n-1}} a^{1} (-1)^{n-1} + {{n}choose{n}} a^{0} (-1)^n

Most of the terms are multiples of a^2. The modulo "removes" them. And keep in mind that a^0 = 1 and {{n}choose{n}} = 1 and {{n}choose{n-1}} = n:
(a+1)^n mod a^2 = na + 1
(a-1)^n mod a^2 = na (-1)^{n-1} + (-1)^n

If n is even then:
((a+1)^n + (a-1)^n) mod a^2 = (na + 1 + na * (-1) + 1) mod a^2 = 2 mod a^2
→ For any even n the result is always 2.

If n is odd then:
((a+1)^n + (a-1)^n) mod a^2 = (na + 1 + na - 1) mod a^2 = 2na mod a^2 = 2n mod a

The maximum n_{max} such that 2n_{max} is as close as possible to a:
2n_{max} = a - 1

n_{max} = dfrac{a - 1}{2}

n_{max} must be an integer, therefore the division is actually an integer division:
n_{max} = \lfloor dfrac{a - 1}{2} \rfloor

And finally the remainder becomes:
r_{max} = 2an_{max} = 2a \lfloor dfrac{a - 1}{2} \rfloor

## Modifications by HackerRank

I wrote very dirty code to solve a few test cases. Some test cases time out, some give just a wrong result.
Not very proud of it !

# Interactive test

This feature is not available for the current problem.

# My code

… was written in C++ 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>

#define ORIGINAL
#ifdef ORIGINAL
int main()
{
unsigned int sum = 0;

// iterate over 3..1000
for (unsigned int a = 3; a <= 1000; a++)
{
unsigned int n_max = (a - 1) / 2;
sum += 2*a*n_max;
}

// print result
std::cout << sum << std::endl;

return 0;
}

#else

#include <deque>

// return (a*b) % modulo
unsigned long long mulmod(unsigned long long a, unsigned long long b, unsigned long long modulo)
{
// (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;
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;
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;
}

int main()
{
const unsigned int minA   = 1;
const unsigned int Modulo = 1000000007;

std::deque<unsigned int> sums2(minA, 0);
std::deque<unsigned int> sums3(minA, 0);

unsigned int tests = 1;
std::cin >> tests;
while (tests--)
{
unsigned int exponent = 2;
unsigned long long maxA = 1000;
std::cin >> maxA >> exponent;

unsigned long long sum = 0;
if (exponent == 2)
{
// closed formula
{
unsigned __int128 half = maxA / 2;
sum  = half * (8*half*half - 3*half - 5) / 3;

if (maxA % 2 == 1)
sum += (maxA*maxA -   maxA) % Modulo;

sum += 2; // maxA == 1
sum %= Modulo;
std::cout << sum << std::endl;
continue;
}

sum = sums2.back();
for (auto a = sums2.size(); a <= maxA; a++)
{
unsigned long long A = a; // make sure that a*a is computed in  64 bit to avoid overflows
unsigned int even = (A*A - 2*A) % Modulo;
unsigned int odd  = (A*A -   A) % Modulo;

// direct evaluation
if (a == 1) // the formulas below fail for a == 1 ...
sum += 2;
else if (a % 2 == 0)
sum += even; // for even numbers
else
sum += odd;  // for odd  numbers

sum %= Modulo;
sums2.push_back(sum);
}

sum = sums2[maxA];
}
else // exponent = 3
{
sum = sums3.back();
for (auto a = sums3.size(); a <= maxA; a++)
{
unsigned __int128 A = a; // make sure that a*a*a is computed in 128 bit to avoid overflows
unsigned int even = (A*A*A - 2*A) % Modulo;
unsigned int odd  = (A*A*A -   A) % Modulo;

if (a == 1) // the formulas below fail for a == 1 ...
sum += 0;
else if (a % 2 == 0)
sum += even; // for even numbers
else
sum += odd;  // for odd  numbers

sum %= Modulo;
sums3.push_back(sum);
}

sum = sums3[maxA];
}

// print result
std::cout << sum << std::endl;
}

return 0;
}

#endif


This solution contains 31 empty lines, 19 comments and 6 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

May 11, 2017 submitted solution

# Hackerrank

My code solves 5 out of 12 test cases (score: 26.67%)

I failed 2 test cases due to wrong answers and 5 because of timeouts

# Difficulty

Project Euler ranks this problem at 25% (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 119 - Digit power sum Disc game prize fund - problem 121 >>
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