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What is the sum of the digits of the number 2^1000?

This is a problem from Project Euler, and this question includes some source code, so consider this your spoiler alert, in case you are interested in solving it yourself. It is discouraged to distribute solutions to the problems, and that isn't what I want. I just need a little nudge and guidance in the right direction, in good faith.
The problem reads as follows:
2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 2^1000?
I understand the premise and math of the problem, but I've only started practicing C# a week ago, so my programming is shaky at best.
I know that int, long and double are hopelessly inadequate for holding the 300+ (base 10) digits of 2^1000 precisely, so some strategy is needed. My strategy was to set a calculation which gets the digits one by one, and hope that the compiler could figure out how to calculate each digit without some error like overflow:
using System;
using System.IO;
using System.Windows.Forms;
namespace euler016
{
class DigitSum
{
// sum all the (base 10) digits of 2^powerOfTwo
[STAThread]
static void Main(string[] args)
{
int powerOfTwo = 1000;
int sum = 0;
// iterate through each (base 10) digit of 2^powerOfTwo, from right to left
for (int digit = 0; Math.Pow(10, digit) < Math.Pow(2, powerOfTwo); digit++)
{
// add next rightmost digit to sum
sum += (int)((Math.Pow(2, powerOfTwo) / Math.Pow(10, digit) % 10));
}
// write output to console, and save solution to clipboard
Console.Write("Power of two: {0} Sum of digits: {1}\n", powerOfTwo, sum);
Clipboard.SetText(sum.ToString());
Console.WriteLine("Answer copied to clipboard. Press any key to exit.");
Console.ReadKey();
}
}
}
It seems to work perfectly for powerOfTwo < 34. My calculator ran out of significant digits above that, so I couldn't test higher powers. But tracing the program, it looks like no overflow is occurring: the number of digits calculated gradually increases as powerOfTwo = 1000 increases, and the sum of digits also (on average) increases with increasing powerOfTwo.
For the actual calculation I am supposed to perform, I get the output:
Power of two: 1000 Sum of digits: 1189
But 1189 isn't the right answer. What is wrong with my program? I am open to any and all constructive criticisms.

For calculating the values of such big numbers you not only need to be a good programmer but also a good mathematician. Here is a hint for you,
there's familiar formula ax = ex ln a , or if you prefer, ax = 10x log a.
More specific to your problem
21000 Find the common (base 10) log of 2, and multiply it by 1000; this is the power of 10. If you get something like 1053.142 (53.142 = log 2 value * 1000) - which you most likely will - then that is 1053 x 100.142; just evaluate 100.142 and you will get a number between 1 and 10; and multiply that by 1053, But this 1053 will not be useful as 53 zero sum will be zero only.
For log calculation in C#
Math.Log(num, base);
For more accuracy you can use, Log and Pow function of Big Integer.
Now rest programming help I believe you can have from your side.

Normal int can't help you with such a large number. Not even long. They are never designed to handle numbers such huge. int can store around 10 digits (exact max: 2,147,483,647) and long for around 19 digits (exact max: 9,223,372,036,854,775,807). However, A quick calculation from built-in Windows calculator tells me 2^1000 is a number of more than 300 digits.
(side note: the exact value can be obtained from int.MAX_VALUE and long.MAX_VALUE respectively)
As you want precise sum of digits, even float or double types won't work because they only store significant digits for few to some tens of digits. (7 digit for float, 15-16 digits for double). Read here for more information about floating point representation, double precision
However, C# provides a built-in arithmetic
BigInteger for arbitrary precision, which should suit your (testing) needs. i.e. can do arithmetic in any number of digits (Theoretically of course. In practice it is limited by memory of your physical machine really, and takes time too depending on your CPU power)
Back to your code, I think the problem is here
Math.Pow(2, powerOfTwo)
This overflows the calculation. Well, not really, but it is the double precision is not precisely representing the actual value of the result, as I said.

A solution without using the BigInteger class is to store each digit in it's own int and then do the multiplication manually.
static void Problem16()
{
int[] digits = new int[350];
//we're doing multiplication so start with a value of 1
digits[0] = 1;
//2^1000 so we'll be multiplying 1000 times
for (int i = 0; i < 1000; i++)
{
//run down the entire array multiplying each digit by 2
for (int j = digits.Length - 2; j >= 0; j--)
{
//multiply
digits[j] *= 2;
//carry
digits[j + 1] += digits[j] / 10;
//reduce
digits[j] %= 10;
}
}
//now just collect the result
long result = 0;
for (int i = 0; i < digits.Length; i++)
{
result += digits[i];
}
Console.WriteLine(result);
Console.ReadKey();
}

I used bitwise shifting to left. Then converting to array and summing its elements. My end result is 1366, Do not forget to add reference to System.Numerics;
BigInteger i = 1;
i = i << 1000;
char[] myBigInt = i.ToString().ToCharArray();
long sum = long.Parse(myBigInt[0].ToString());
for (int a = 0; a < myBigInt.Length - 1; a++)
{
sum += long.Parse(myBigInt[a + 1].ToString());
}
Console.WriteLine(sum);

since the question is c# specific using a bigInt might do the job. in java and python too it works but in languages like c and c++ where the facility is not available you have to take a array and do multiplication. take a big digit in array and multiply it with 2. that would be simple and will help in improving your logical skill. and coming to project Euler. there is a problem in which you have to find 100! you might want to apply the same logic for that too.

Try using BigInteger type , 2^100 will end up to a a very large number for even double to handle.

BigInteger bi= new BigInteger("2");
bi=bi.pow(1000);
// System.out.println("Val:"+bi.toString());
String stringArr[]=bi.toString().split("");
int sum=0;
for (String string : stringArr)
{ if(!string.isEmpty()) sum+=Integer.parseInt(string); }
System.out.println("Sum:"+sum);
------------------------------------------------------------------------
output :=> Sum:1366

Here's my solution in JavaScript
(function (exponent) {
const num = BigInt(Math.pow(2, exponent))
let arr = num.toString().split('')
arr.slice(arr.length - 1)
const result = arr.reduce((r,c)=> parseInt(r)+parseInt(c))
console.log(result)
})(1000)

This is not a serious answer—just an observation.
Although it is a good challenge to try to beat Project Euler using only one programming language, I believe the site aims to further the horizons of all programmers who attempt it. In other words, consider using a different programming language.
A Common Lisp solution to the problem could be as simple as
(defun sum_digits (x)
(if (= x 0)
0
(+ (mod x 10) (sum_digits (truncate (/ x 10))))))
(print (sum_digits (expt 2 1000)))

main()
{
char c[60];
int k=0;
while(k<=59)
{
c[k]='0';
k++;
}
c[59]='2';
int n=1;
while(n<=999)
{
k=0;
while(k<=59)
{
c[k]=(c[k]*2)-48;
k++;
}
k=0;
while(k<=59)
{
if(c[k]>57){ c[k-1]+=1;c[k]-=10; }
k++;
}
if(c[0]>57)
{
k=0;
while(k<=59)
{
c[k]=c[k]/2;
k++;
}
printf("%s",c);
exit(0);
}
n++;
}
printf("%s",c);
}

Python makes it very simple to compute this with an oneliner:
print sum(int(digit) for digit in str(2**1000))
or alternatively with map:
print sum(map(int,str(2**1000)))

Fermat primality test

I have tried to write a code for Fermat primality test, but apparently failed.
So if I understood well: if p is prime then ((a^p)-a)%p=0 where p%a!=0.
My code seems to be OK, therefore most likely I misunderstood the basics. What am I missing here?
private bool IsPrime(int candidate)
{
//checking if candidate = 0 || 1 || 2
int a = candidate + 1; //candidate can't be divisor of candidate+1
if ((Math.Pow(a, candidate) - a) % candidate == 0) return true;
return false;
}

Reading the wikipedia article on the Fermat primality test, You must choose an a that is less than the candidate you are testing, not more.
Furthermore, as MattW commented, testing only a single a won't give you a conclusive answer as to whether the candidate is prime. You must test many possible as before you can decide that a number is probably prime. And even then, some numbers may appear to be prime but actually be composite.

Your basic algorithm is correct, though you will have to use a larger data type than int if you want to do this for non-trivial numbers.
You should not implement the modular exponentiation in the way that you did, because the intermediate result is huge. Here is the square-and-multiply algorithm for modular exponentiation:
function powerMod(b, e, m)
x := 1
while e > 0
if e%2 == 1
x, e := (x*b)%m, e-1
else b, e := (b*b)%m, e//2
return x
As an example, 437^13 (mod 1741) = 819. If you use the algorithm shown above, no intermediate result will be greater than 1740 * 1740 = 3027600. But if you perform the exponentiation first, the intermediate result of 437^13 is 21196232792890476235164446315006597, which you probably want to avoid.
Even with all of that, the Fermat test is imperfect. There are some composite numbers, the Carmichael numbers, that will always report prime no matter what witness you choose. Look for the Miller-Rabin test if you want something that will work better. I modestly recommend this essay on Programming with Prime Numbers at my blog.

You are dealing with very large numbers, and trying to store them in doubles, which is only 64 bits.
The double will do the best it can to hold your number, but you are going to loose some accuracy.
An alternative approach:
Remember that the mod operator can be applied multiple times, and still give the same result.
So, to avoid getting massive numbers you could apply the mod operator during the calculation of your power.
Something like:
private bool IsPrime(int candidate)
{
//checking if candidate = 0 || 1 || 2
int a = candidate - 1; //candidate can't be divisor of candidate - 1
int result = 1;
for(int i = 0; i < candidate; i++)
{
result = result * a;
//Notice that without the following line,
//this method is essentially the same as your own.
//All this line does is keeps the numbers small and manageable.
result = result % candidate;
}
result -= a;
return result == 0;
}

Most probable bits in random integer

I've made such experiment - made 10 million random numbers from C and C#. And then counted how much times each bit from 15 bits in random integer is set. (I chose 15 bits because C supports random integer only up to 0x7fff).
What i've got is this:
I have two questions:
Why there are 3 most probable bits ? In C case bits 8,10,12 are most probable. And
in C# bits 6,8,11 are most probable.
Also seems that C# most probable bits is mostly shifted by 2 positions then compared to C most probable bits. Why is this ? Because C# uses other RAND_MAX constant or what ?
My test code for C:
void accumulateResults(int random, int bitSet[15]) {
int i;
int isBitSet;
for (i=0; i < 15; i++) {
isBitSet = ((random & (1<<i)) != 0);
bitSet[i] += isBitSet;
}
}
int main() {
int i;
int bitSet[15] = {0};
int times = 10000000;
srand(0);
for (i=0; i < times; i++) {
accumulateResults(rand(), bitSet);
}
for (i=0; i < 15; i++) {
printf("%d : %d\n", i , bitSet[i]);
}
system("pause");
return 0;
}
And test code for C#:
static void accumulateResults(int random, int[] bitSet)
{
int i;
int isBitSet;
for (i = 0; i < 15; i++)
{
isBitSet = ((random & (1 << i)) != 0) ? 1 : 0;
bitSet[i] += isBitSet;
}
}
static void Main(string[] args)
{
int i;
int[] bitSet = new int[15];
int times = 10000000;
Random r = new Random();
for (i = 0; i < times; i++)
{
accumulateResults(r.Next(), bitSet);
}
for (i = 0; i < 15; i++)
{
Console.WriteLine("{0} : {1}", i, bitSet[i]);
}
Console.ReadKey();
}
Very thanks !! Btw, OS is Windows 7, 64-bit architecture & Visual Studio 2010.
EDIT
Very thanks to #David Heffernan. I made several mistakes here:
Seed in C and C# programs was different (C was using zero and C# - current time).
I didn't tried experiment with different values of Times variable to research reproducibility of results.
Here's what i've got when analyzed how probability that first bit is set depends on number of times random() was called:
So as many noticed - results are not reproducible and shouldn't be taken seriously.
(Except as some form of confirmation that C/C# PRNG are good enough :-) ).

This is just common or garden sampling variation.
Imagine an experiment where you toss a coin ten times, repeatedly. You would not expect to get five heads every single time. That's down to sampling variation.
In just the same way, your experiment will be subject to sampling variation. Each bit follows the same statistical distribution. But sampling variation means that you would not expect an exact 50/50 split between 0 and 1.
Now, your plot is misleading you into thinking the variation is somehow significant or carries meaning. You'd get a much better understanding of this if you plotted the Y axis of the graph starting at 0. That graph looks like this:
If the RNG behaves as it should, then each bit will follow the binomial distribution with probability 0.5. This distribution has variance np(1 − p). For your experiment this gives a variance of 2.5 million. Take the square root to get the standard deviation of around 1,500. So you can see simply from inspecting your results, that the variation you see is not obviously out of the ordinary. You have 15 samples and none are more than 1.6 standard deviations from the true mean. That's nothing to worry about.
You have attempted to discern trends in the results. You have said that there are "3 most probable bits". That's only your particular interpretation of this sample. Try running your programs again with different seeds for your RNGs and you will have graphs that look a little different. They will still have the same quality to them. Some bits are set more than others. But there won't be any discernible patterns, and when you plot them on a graph that includes 0, you will see horizontal lines.
For example, here's what your C program outputs for a random seed of 98723498734.
I think this should be enough to persuade you to run some more trials. When you do so you will see that there are no special bits that are given favoured treatment.

You know that the deviation is about 2500/5,000,000, which comes down to 0,05%?

Note that the difference of frequency of each bit varies by only about 0.08% (-0.03% to +0.05%). I don't think I would consider that significant. If every bit were exactly equally probable, I would find the PRNG very questionable instead of just somewhat questionable. You should expect some level of variance in processes that are supposed to be more or less modelling randomness...

Longest recurring cycle in its decimal fraction - a bug or a misunderstanding?

This is fairly 'math-y' but I'm posting here because it's a Project Euler problem, & I have working code that presumably has bugs in it.
The question Determing longest repeating cycle in a decimal expansion solves the problem using logarithms, but I'm interested in solving with simple brute force. More accurately, I'm interested in understanding why my algorithm and code is not returning the correct solution.
The algorithm is simple:
replicate a 'long division',
at each step record the divisor and the remainder
when a divisor / remainder tuple is repeated, infer that the decimal representation will repeat.
Here are private fields, as requested
private int numerator;
private int recurrence;
private int result;
private int resultRecurrence;
private List<dynamic> digits;
and here is the code:
private void Go()
{
foreach (var i in primes)
{
digits = new List<dynamic>();
numerator = 1;
recurrence = 0;
while (numerator != 0)
{
numerator *= 10;
// quotient
var q = numerator / i;
// remainder
var r = numerator % i;
digits.Add(new { Divisor = q, Remainder = r });
// if we've found a repetition then break out
var m = digits.Where(p => p.Divisor == q && p.Remainder == r).ToList();
if (m.Count > 1)
{
recurrence = digits.LastIndexOf(m[0]) - digits.IndexOf(m[0]);
break;
}
numerator = r;
}
if (recurrence > resultRecurrence)
{
resultRecurrence = recurrence;
result = i;
}
}}
When testing integers < 10 and < 20 I get the correct result; and I correctly identify the value of i as well. However the decimal represetation that I get is incorrect - I calculate i-1 whereas the correct result is far less (something like i-250).
So presumably I either have a programming bug - which I can't find - or a logic bug.
I'm confused because it feels like a multiplicative group over p to me, in which there would be p-1 elements. I'm sure I'm missing something, can anyone provide suggestions?
edit
I'm not going to include my prime number code - it's not relevant, as I explain above I correctly identify the value of i (from memory it is 983) but I'm having problems getting the correct value for resultRecurrence.

I'm confused because it feels like a multiplicative group over p to me, in which there would be p-1 elements. I'm sure I'm missing something, can anyone provide suggestions?
Close.
For all primes except 2 and 5 (which divide 10), the sequence of remainders is formed by starting with 1 and transforming by
remainder = (10 * remainder) % prime
thus the k-th remainder is 10k (mod prime) and the set of remainders forms a subgroup of the group of nonzero remainders modulo prime[1]. The length of the recurring cycle is the order of that subgroup, which is also known as the order of 10 modulo prime.
The order of the group of nonzero remainders modulo prime is prime-1, and there's a theorem by Fermat:
Let G be a finite group of order g and H be a subgroup of G. Then the order h of H divides g.
So the length of the cycle is always a divisor of prime-1, and sometimes it's prime-1, e.g. for 7 or 19.
[1] For composite numbers n coprime to 10, that would be the group of remainders modulo n that are coprime to n.

First off, you don’t need the divisors, you only need the remainders.
Secondly, I would split the function into multiple independent parts instead of having everything in one big method: The long division / finding of the cycle length is independent of the rest (= finding the longest cycle).
Your break on Where coupled with Count is unintuitive. Why not just use a while loop with the condition (! digits.Contains(r))? (This would require putting 0 as a remainder into the digits list before the loop start.)
This leaves us with a much cleaner code that should be straightforward to debug.

recurrence = digits.LastIndexOf(m[0]) - digits.IndexOf(m[0]);
Surely the value of resultRecurrence is always going to be i-1 ? Since for a fraction of the form 1/n, the decimal starts repeating exactly when the division-in-progress (the ith digit) gives the same quotient-remainder as the very first trial division (1, hence i-1).
(as a side note, may I introduce you to Math.DivRem).

Series calculation

I have some random integers like
99 20 30 1 100 400 5 10
I have to find a sum from any combination of these integers that is closest(equal or more but not less) to a given number like
183
what is the fastest and accurate way of doing this?

If your numbers are small, you can use a simple Dynamic Programming(DP) technique. Don't let this name scare you. The technique is fairly understandable. Basically you break the larger problem into subproblems.
Here we define the problem to be can[number]. If the number can be constructed from the integers in your file, then can[number] is true, otherwise it is false. It is obvious that 0 is constructable by not using any numbers at all, so can[0] is true. Now you try to use every number from the input file. We try to see if the sum j is achievable. If an already achieved sum + current number we try == j, then j is clearly achievable. If you want to keep track of what numbers made a particular sum, use an additional prev array, which stores the last used number to make the sum. See the code below for an implementation of this idea:
int UPPER_BOUND = number1 + number2 + ... + numbern //The largest number you can construct
bool can[UPPER_BOUND + 1]; //can[number] is true if number can be constructed
can[0] = true; //0 is achievable always by not using any number
int prev[UPPER_BOUND + 1]; //prev[number] is the last number used to achieve sum "number"
for (int i = 0; i < N; i++) //Try to use every number(numbers[i]) from the input file
{
for (int j = UPPER_BOUND; j >= 1; j--) //Try to see if j is an achievable sum
{
if (can[j]) continue; //It is already an achieved sum, so go to the next j
if (j - numbers[i] >= 0 && can[j - numbers[i]]) //If an (already achievable sum) + (numbers[i]) == j, then j is obviously achievable
{
can[j] = true;
prev[j] = numbers[i]; //To achieve j we used numbers[i]
}
}
}
int CLOSEST_SUM = -1;
for (int i = SUM; i <= UPPER_BOUND; i++)
if (can[i])
{
//the closest number to SUM(larger than SUM) is i
CLOSEST_SUM = i;
break;
}
int currentSum = CLOSEST_SUM;
do
{
int usedNumber = prev[currentSum];
Console.WriteLine(usedNumber);
currentSum -= usedNumber;
} while (currentSum > 0);

This seems to be a Knapsack-like problem, where the value of your integers would be the "weight" of each item, the "profit" of each item is 1, and you are looking for the least number of items to exactly sum to the maximum allowable weight of the knapsack.

This is a variant of the SUBSET-SUM problem, and is also NP-Hard like SUBSET-SUM.
But if the numbers involved are small, pseudo-polynomial time algorithms exist. Check out:
http://en.wikipedia.org/wiki/Subset_sum_problem
Ok More details.
The following problem:
Given an array of integers, and integers a,b, is there
some subset whose sum lies in the
interval [a,b] is NP-Hard.
This is so because we can solve subset-sum by choosing a=b=0.
Now this problem easily reduces to your problem and so your problem is NP-Hard too.
Now you can use the polynomial time approximation algorithm mentioned in the wiki link above.
Given an array of N integers, a target S and an approximation threshold c,
there is a polynomial time approximation algorithm (involving 1/c) which tells if there is a subset sum in the interval [(1-c)S, S].
You can use this repeatedly (by some form of binary search) to find the best approximation to S you need. Note you can also use this on intervals of the from [S, (1+c)S], while the knapsack will only give you a solution <= S.
Of course there might be better algorithms, in fact I can bet on it. There should be plenty of literature on the web. Some search terms you can use: approximation algorithms for subset-sum, pseudo-polynomial time algorithms, dynamic programming algorithm etc.

A simple-brute-force-method would be to read the text in, parse it into numbers, and then go through all combinations until you find the required sum.
A quicker solution would be to sort the numbers, then...
Add the largest number to your sum, Is it too big? if so, take it off and try the next smallest.
if the sum is too small, add the next largest number and repeat.
Continue adding numbers not letting the sum exceed the target. Finish when you hit the target.
Note that when you backtrack, you may need to back track more than one level. Sounds like a good case for recursion...

If the numbers are large you can turn this into an Integer Programme. Using Mathematicas solver, it might look something like this
nums = {99, 20, 30 , 1, 100, 400, 5, 10};
vars = a /# Range#Length#nums;
Minimize[(vars.nums - 183)^2, vars, Integers]

You can sort the list of values, find the first value that's greater than the target, and start concentrating on the values that are less than the target. Find the sum that's closest to the target without going over, then compare that to the first value greater than the target. If the difference between the closest sum and the target is less than the difference between the first value greater than the target and the target, then you have the sum that's closest.
Kinda hokey, but I think the logic hangs together.