1. 1. Multiples of 3 and 5

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

233168
sum(i for i in range(1000) if i%3 == 0 or i%5 == 0)

2. 2. Even Fibonacci numbers

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.

4613732
def fib(max):
    a,b = 1,2
    while a <= max:
        yield a
        a,b = b,a+b
sum(i for i in fib(4000000) if i%2 == 0)

3. 3. Largest prime factor

The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

6857
def prime(max):
    primes = [2]

    for test in range(primes[-1]+1, max+1, 2):
        sqrt = int(test ** 0.5)
        isprime = True
        for i in primes:
            if i > sqrt:
                break
            if test % i == 0:
                isprime = False
                break
        if isprime:
            primes.append(test)

    for i in primes:
        yield i

num = 600851475143 

for i in prime(int(num ** 0.5)):
    if num % i == 0:
        max = i

print(max)

4. 4. Largest palindrome product

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.

906609
def reverse(num):
    renum = 0
    while num:
        renum = renum * 10 + num % 10
        num //= 10
    return renum

def ispalindrome(num):
    return num == reverse(num)

max(i*j for i in range(999) for j in range(999) if ispalindrome(i*j))

5. 5. Smallest multiple

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?

232792560
def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a%b)

def GCD(a, b):
    while b:
        a, b = b, a%b
    return a

def lcm(m,n):
    return m*n//gcd(m,n)

def smallest_multiple(n):
    if n == 1:
        return 1
    return lcm(n, smallest_multiple(n-1))

smallest_multiple(20)

6. 6. Sum square difference

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.

25164150
sum(range(101))** 2 - sum(x**2 for x in range(101))

7. 7. 10001st prime

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?

104743
def prime():
    primes = [2]
    x = 1
    while True:
        x += 2
        sqrt = int(x ** 0.5)
        isprime = True
        for i in primes:
            if i > sqrt:
                break
            if x % i == 0:
                isprime = False
                break
        if isprime:
            yield primes[-1]
            primes.append(x)
def primeByIndex(n):
    p = prime()
    x = 0
    for i in range(n):
        x = next(p)
    return x
primeByIndex(10001)

8. 8. Largest product in a series

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

40824
d = '''73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450'''

num = 5
dd = '1'*num + d.replace('\n','').replace('0','1'*num)

product = 1
max_product = 0

for i in range(len(dd)-num):
    product = product * int(dd[i+num]) // int(dd[i])
    max_product = max(product, max_product)

9. 9. Special Pythagorean triplet

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.

----> a < 333 && c > 333

200 375 425
31875000

for a in range(1, 333):
    for c in range(333, 1000):
        b = 1000-a-c
        if a*a + b*b == c*c:
            print(a, b, c)
            print(a*b*c)
            break

10. 10. Summation of primes

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

Find the sum of all the primes below two million.

142913828922
def prime(limit=0):
    primes = [2, 3]
    for i in primes:
        yield i

    def is_prime(test):
        sqrt = int(test ** 0.5)
        isprime = True
        for i in primes:
            if i > sqrt:
                break
            if test % i == 0:
                isprime = False
                break
        return isprime

    t = 0
    while True:
        t += 6
        if limit and t > limit:
            break
        for x in [t-1, t+1]:
            if is_prime(x):
                primes.append(x)
                yield x

sum(p for p in prime(2000000))

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