I've copied the whole given table to a spreadsheet (open office was good enough). Then, I've summed them all and divided by 10^42 to get the first 10 digits.
Here are the first leading 10 digits: 5537376230.
ProjectEuler solver
Friday, April 15, 2011
Project Euler - Problem 10 - C#
http://projecteuler.net/project/resources/010_7c4950764b52402fe1d29323af4e6c6f/010_overview.pdf
//The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. //Find the sum of all the primes below two million. static UInt64 Problem10() { UInt64 sum_of_primes = 0; foreach (var prime in primes(2000000)) { sum_of_primes += prime; //Console.Write(prime.ToString() + ", "); } return sum_of_primes; // Problem10: 142913828922 } static IEnumerable<UInt64> primes(UInt64 limit) { List<UInt64> primes_list = new List<UInt64>(); // reset with 0 bool[] nums_list = new bool[limit]; // reset with 0 /* Create a list of all candidates. * false - Marks an invalid candidate. * true - A valid candidate. */ nums_list[2] = true; primes_list.Add(2); for (UInt64 current_num = 3; current_num <= limit; current_num += 2) { nums_list[current_num] = true; } // Sieve of Eratosthenes algorithm for (UInt64 current_num = 3; current_num <= limit; current_num += 2) { if (nums_list[current_num] == true) { primes_list.Add(current_num); for (UInt64 elemenator = 2 * current_num; elemenator < limit; elemenator += current_num) { nums_list[elemenator] = false; } } } foreach (var prime in primes_list) { yield return prime; } }
Project Euler - Problem 9 - C#
/* A Pythagorean triplet is a set of three natural numbers, a b c, for which, a^2 + b^2 = c^2 For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2. There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc. */ static int Problem9() { /* HLD: * a + b + c = 1000 ==> c = 1000 - a - b * foreach {a, b} in {{1,1}, {1,2} ... {2,2} ... {499,1} ... {499,499}} * predicate of pass: a^2 + b^2 = c^2 */ var solution_found = false; int a = 0, b = 0; int c = 1000; for (b = 1; b < 499; b++) { c = 1000; for (a = b; a <= c; a++) { c = 1000 - a - b; if ((a * a + b * b) == (c * c)) { solution_found = true; break; } } if (solution_found) { break; } } return a * b * c; // Problem9: 31875000 }
Thursday, April 14, 2011
Project Euler - Problem 8 - C#
/*
Find the greatest product of five consecutive digits in the 1000-digit number.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
*/
static int Problem8()
{
StringBuilder data = new StringBuilder(1000);
data.Append("73167176531330624919225119674426574742355349194934");
data.Append("96983520312774506326239578318016984801869478851843");
data.Append("85861560789112949495459501737958331952853208805511");
data.Append("12540698747158523863050715693290963295227443043557");
data.Append("66896648950445244523161731856403098711121722383113");
data.Append("62229893423380308135336276614282806444486645238749");
data.Append("30358907296290491560440772390713810515859307960866");
data.Append("70172427121883998797908792274921901699720888093776");
data.Append("65727333001053367881220235421809751254540594752243");
data.Append("52584907711670556013604839586446706324415722155397");
data.Append("53697817977846174064955149290862569321978468622482");
data.Append("83972241375657056057490261407972968652414535100474");
data.Append("82166370484403199890008895243450658541227588666881");
data.Append("16427171479924442928230863465674813919123162824586");
data.Append("17866458359124566529476545682848912883142607690042");
data.Append("24219022671055626321111109370544217506941658960408");
data.Append("07198403850962455444362981230987879927244284909188");
data.Append("84580156166097919133875499200524063689912560717606");
data.Append("05886116467109405077541002256983155200055935729725");
data.Append("71636269561882670428252483600823257530420752963450");
const int aggregation_consecutives = 5;
// Starting point initialization
var data_string = data.ToString();
var global_max_mul = 0;
var consicutives_so_far = 0;
var local_max_mul = 1;
var tail_num = 0;
var current_num = 0;
for (int current_char_idx = 0; current_char_idx < data.Length; current_char_idx++)
{
current_num = (int)data_string[current_char_idx] - '0';
if (current_num == 0)
{
consicutives_so_far = 0;
local_max_mul = 1;
continue;
}
if (consicutives_so_far < aggregation_consecutives)
{
local_max_mul *= current_num;
global_max_mul = Math.Max(global_max_mul, local_max_mul);
consicutives_so_far++;
continue;
}
tail_num = data_string[current_char_idx - 5] - '0';
local_max_mul = (local_max_mul / tail_num) * current_num;
global_max_mul = Math.Max(global_max_mul, local_max_mul);
consicutives_so_far++;
Console.WriteLine(string.Format(
"current_char_idx:{0}, consicutives_so_far={1}, current_num={2} local_max_mul={3}",
current_char_idx, consicutives_so_far, current_num, local_max_mul));
}
return global_max_mul;
// Problem8: 40824
}
Wednesday, April 13, 2011
Project Euler - Problem 7 - Tcl
#By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. #What is the 10001st prime number? proc problem7-v2 {} { # Much faster. #REMARK: Based on problem-3 set primes [list 2 3] set current_number 5 set primes_found 2 set finished FALSE while { !$finished } { set is_prime TRUE foreach prime $primes { if { ( $current_number % $prime ) == 0 } { set is_prime FALSE break } if { $prime > ($current_number / 2) } { # Can't be bigger than factor 2! break } } if { $is_prime } { incr primes_found 1 set last_prime $current_number lappend primes $last_prime } set finished [expr $primes_found == 10001] incr current_number 2 } puts "solution-7: [lindex $primes end]" } # solution-7: solution-7: 104743
Project Euler - Problem 6 - Tcl
#The sum of the squares of the first ten natural numbers is, #1^2 + 2^2 + ... + 10^2 = 385 # #The square of the sum of the first ten natural numbers is, #(1 + 2 + ... + 10)^2 = 55^2 = 3025 # #Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 - 385 = 2640. # #Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum. proc problem6 {} { set sum_of_squares 0 set squares_of_sum 0 for { set idx 1 } { $idx <= 100 } {incr idx} { set sum_of_squares [expr $sum_of_squares + ($idx * $idx)] set squares_of_sum [expr $squares_of_sum + $idx ] } set squares_of_sum [expr $squares_of_sum*$squares_of_sum] puts "Solution-6: [expr $squares_of_sum - $sum_of_squares]" } #Solution-6: 25164150
Project Euler - Problem 4 - Tcl
#A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 99. #Find the largest palindrome made from the product of two 3-digit numbers. proc problem4 {} { set palindrom_found FALSE set current_palindrom [expr 999*999] while { $palindrom_found == FALSE } { set current_palindrom [next_polindrom $current_palindrom -1] for { set idx 999 } { $idx > 900 } {incr idx -1} { if { ( $current_palindrom % $idx ) == 0 && ($current_palindrom / $idx) < 1000 } { puts "$idx*[expr $current_palindrom / $idx]=$current_palindrom" set palindrom_found TRUE break } } } } #****f* math/next_polindrom # FUNCTION # Generates the next polindrom according to a given polindrom. # # INPUTS # pre_polindrom: # * TBD # direction: # * (+n) - next (n) polindrom up # * (-n) - next (n) polindrom down # # RESULT # * int - The consecutive (n) polinom from the given number/polinom. # * {} - error with parameters or none found. # # EXAMPLE # TBD # # NOTES # * Good for solving problem-4. Jumps from 100000 to 9999 (see BUGS). # # BUGS # * Wrong sequance: {... 102201 101101 100001 9999 9889 9779 ... } # # SYNOPSIS proc next_polindrom { pre_polindrom direction } { # SOURCE # E.g. pre_polindrom ==> 123000 set polindrom_len [expr round( ceil( log10( $pre_polindrom ))) ] set polindrom_mid_len [expr ($polindrom_len / 2) + ($polindrom_len % 2)] set digits_to_align [expr $polindrom_len - $polindrom_mid_len] set polindrom_core $pre_polindrom ; # E.g. polindrom_core ==> 123000 while { $digits_to_align > 0} { set polindrom_core [expr $polindrom_core / 10] incr digits_to_align -1 } ; # E.g. polindrom_core ==> 123 incr polindrom_core $direction ; # E.g. (of +1): polindrom_core ==> 124 set splited_polindrom_core [split $polindrom_core {}] ; # E.g. splited_polindrom_core ==> {1 2 3} set splited_reversed_polindrom_core [lreverse $splited_polindrom_core] ; # E.g. splited_reversed_polindrom_core ==> {3 2 1} set splited_reversed_polindrom_core [lrange $splited_reversed_polindrom_core [expr $polindrom_len % 2] end] ; # E.g. for 123000: {3 2 1} ==> {3 2 1}, for 12300: {3 2 1} ==> {2 1} set reversed_polindrom_core [join $splited_reversed_polindrom_core {}] ; # E.g. splited_reversed_polindrom_core ==> {321} set next_polinom "$polindrom_core$reversed_polindrom_core" return $next_polinom ###### } #Solution-4: 993*913=906609
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