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[professorlamp]

Evenin,

I thought I'd just post up the questions and answers (as I work my way through) to some of the project euler problems. Feel free to hand in your own answers and hopefully we can have a single source for all the answers. Right now all of the answers are just scattered across the internet and are in various languages.

Note: All of the solutions I give are written in Python 3.2, I am also a n00b (hence why I'm doing this) so it might take me a while to post some solutions unless someone else gets their first.
Note: Project Eulers site  http://projecteuler.net/

CODERS COME FORTH

Spoiler: Question 1 (click to show/hide)

#1 If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.Find the sum of all the multiples of 3 or 5 below 1000.

Spoiler: Solution 1 (click to show/hide)

Code: [Select]

while a < 1000:
    if a%3 == 0 or a%5 == 0:
        print (a)
        asum += a
    a+=1
print (asum)

Spoiler: Question2 (click to show/hide)

#2 Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

Spoiler: Solution2 (click to show/hide)

Code: [Select]

a1 = 1
a2 = 1
a3 = 0
evenSum = 0

while a3 < 4000000:
    a3 = a1 + a2
    a1 = a2
    a2 = a3
    if a3%2 == 0:
        evenSum += a3
print evenSum

Spoiler: Question3 (click to show/hide)

 The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ?

Spoiler: Solution3 (click to show/hide)

Code: [Select]

number = 600851475143
factors = []
d = 2

while number > 1:
    while number%d == 0:
        factors.append(d)
        number /= d
    d += 1

print factors[-1]

Spoiler: Question4 (click to show/hide)

 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.

Spoiler: Solution4 (click to show/hide)

Code: [Select]

number = 999999 # 999^2
palindromic = False

fact1 = 1000
fact2 = 1000
factored = False

while not palindromic  and number >= 100000:
    numstring = str(number)

    if (numstring[0] == numstring[5]) and (numstring[1] == numstring[4])and (numstring[2] == numstring[3]):
        palindromic = True
        print number

    if palindromic:
        for i in range(fact1):
            for k in range(fact2):
                if i*k == number:
                    print i,k
                    factored = True       
    if not factored:
        palindromic = False
        number -= 1

print number

Spoiler: Question5 (click to show/hide)

 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

Spoiler: Solution5 (click to show/hide)

Code: [Select]

#Written in Python 2.7.3

number = 0;
factor = 2;
factored = False;

while (not factored):
    number += 20; #Adding by the largest factor is much faster than adding by 1.
    while (number%factor == 0):
        factor += 1;
    if (factor > 20):
        factored = True;
    else:
        factor = 2;

print number;

Spoiler: Solution5(JAVA) (click to show/hide)

Code: [Select]

class Euler5 {

//Written in Java

static long gcd(long a, long b) {

//Euclid's algorithm
if (b == 0) return a;
else return gcd(b, a%b);
}

static long lcm(long a, long b) {

if (a == 0 || b == 0) return 0;
return a*b/gcd(a,b);

}

static long fac(long a) {
if (a == 1 || a == 0) return 1;
else return a*fac(a-1);
}

public static void main(String args[]) {

long x = 2;

for(int i = 3; i < 21; i++) {
x = lcm(x, i);
}
System.out.println("Least multiple: " + x);
System.out.println("Compare 20!: " + fac(20));


//Verification
for(int i = 2; i< 21; i++) {
if (x%i != 0) System.out.println(i + " doesn't divide " + x);
}
}
}

Spoiler: Question6 (click to show/hide)

 The sum of the squares of the first ten natural numbers is,
12 + 22 + ... + 102 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)2 = 552 = 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.

Spoiler: Solution6 (click to show/hide)

Code: [Select]

#Written in Python 2.7.3

sum_of_squares = 0;
square_of_sum = 0;
counter = 1;

while (counter <= 100):
    sum_of_squares += (counter*counter);
    counter += 1;

counter = 1;

while (counter <= 100):
    square_of_sum += counter;
    counter += 1;

square_of_sum *= square_of_sum;

print square_of_sum - sum_of_squares;

Spoiler: Question7 (click to show/hide)

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 10,001st prime number?

Spoiler: Question8 (click to show/hide)

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

Spoiler: Question9 (click to show/hide)

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a2 + b2 = c2
For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

Spoiler: Question10 (click to show/hide)

 The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.

[da dwarf lord]

Here's my solution for problem 2

Spoiler (click to show/hide)

Code: [Select]

a1 = 1
a2 = 1
a3 = 0
evenSum = 0

while a3 < 4000000:
    a3 = a1 + a2
    a1 = a2
    a2 = a3
    if a3%2 == 0:
        evenSum += a3

print evenSum

Didn't know this existed till is saw this thread, thanks 

EDIT:

Spoiler: Question 3 (click to show/hide)

Code: [Select]

number = 600851475143
factors = []
d = 2

while number > 1:
    while number%d == 0:
        factors.append(d)
        number /= d
    d += 1

print factors[-1]
       

Spoiler: Question 4 (click to show/hide)

Code: [Select]

number = 999999 # 999^2
palindromic = False

fact1 = 1000
fact2 = 1000
factored = False

while not palindromic  and number >= 100000:
    numstring = str(number)

    if (numstring[0] == numstring[5]) and (numstring[1] == numstring[4])and (numstring[2] == numstring[3]):
        palindromic = True
        print number

    if palindromic:
        for i in range(fact1):
            for k in range(fact2):
                if i*k == number:
                    print i,k
                    factored = True       
    if not factored:
        palindromic = False
        number -= 1

print number


[Killjoy]

Oh I remember I had a lot of fun solving the first 50 or so questions back when I was starting computer science. Then I kind of got bored of it. They are not really programming challenges, besides question 18, 67 (As far as I know) where you can feel a little bit of CS.

I solved them all having a python console open, but I see people solving these using pen, paper and their brain filled with smarts I will never hope to have.

Either way, have fun.

[P(ony)SI]

Spoiler: Question 5 (click to show/hide)

Code: [Select]

#Written in Python 2.7.3

number = 0;
factor = 2;
factored = False;

while (not factored):
    number += 20; #Adding by the largest factor is much faster than adding by 1.
    while (number%factor == 0):
        factor += 1;
    if (factor > 20):
        factored = True;
    else:
        factor = 2;

print number;

Spoiler: Question 6 (click to show/hide)

Code: [Select]

#Written in Python 2.7.3

sum_of_squares = 0;
square_of_sum = 0;
counter = 1;

while (counter <= 100):
    sum_of_squares += (counter*counter);
    counter += 1;

counter = 1;

while (counter <= 100):
    square_of_sum += counter;
    counter += 1;

square_of_sum *= square_of_sum;

print square_of_sum - sum_of_squares;

[da_nang]

Spoiler: Question 5 (click to show/hide)

Code: [Select]

class Euler5 {

//Written in Java

static long gcd(long a, long b) {

//Euclid's algorithm
if (b == 0) return a;
else return gcd(b, a%b);
}

static long lcm(long a, long b) {

if (a == 0 || b == 0) return 0;
return a*b/gcd(a,b);

}

static long fac(long a) {
if (a == 1 || a == 0) return 1;
else return a*fac(a-1);
}

public static void main(String args[]) {

long x = 2;

for(int i = 3; i < 21; i++) {
x = lcm(x, i);
}
System.out.println("Least multiple: " + x);
System.out.println("Compare 20!: " + fac(20));


//Verification
for(int i = 2; i< 21; i++) {
if (x%i != 0) System.out.println(i + " doesn't divide " + x);
}
}
}