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 number that is evenly divisible by all of the numbers from 1 to 20?

Analysis

Just find the least common multiple for the integers {2, 3, 4, …, 20} using Euclid’s algorithm.

Solution

Runs < 1 second in Python.

def gcd(a,b): 
  return b and gcd(b, a % b) or a
 
def lcm(a,b): 
  return a * b / gcd(a,b)
 
n = 1
for i in range(2, 21):
    n = lcm(n, i)
print "Answer to PE5 = ",n

Comments

The sum of the squares of the first ten natural numbers is,

The square of the sum of the first ten natural numbers is,

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.

Analysis

The formula for finding the square of the sum is adapted from Problem 1: (n*(n+1)/2)**2. The formula for finding the sum of the squares is: n * (n+1) * (2*n+1) / 6. We use these formulas and find the difference between the sum of the squares and the square of the sum.

Solution

Runs < 1 second in Python.

n = 100;
sqofsum = (n*(n+1) / 2) ** 2
sumofsq = n * (n+1) * (2*n+1) / 6
print "Answer to PE6 = ",sqofsum - sumofsq

Comments

Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
… {continued}

Analysis

Starting with the largest 5 digit number, 99999, count downwards until that number is a substring of the 1000 digit number. Split out the five digits and multiply them together. Ignore any products that could evaluate to zero.

Solution

Runs < 1 second in Perl.

$string = <DATA>;
$p=1;
 
for ($i=99999; ; $i--) {
  if ($i !~ /0/ && $string =~ /$i/) {
    $p*=$_ for split //,$i;
    print "Answer to PE8 = $p";
    exit;
  }
}
__DATA__
731671765313306249...

Comments

Finds the maximum in 121 iterations.

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.

Analysis

The most efficient way to find the largest palindrome product of two 3-digit numbers is by searching odd factors downwards from 999. We test each product as a palindrome and save it. Our search ends when the next product falls below the largest palindrome found.

For a product to qualify as a palindrome the first and last digit must be the same. In this case it will always be 9, an odd number that can only be derived by the multiplication of two odd numbers.

Solution

Runs < 1 second in Python.

def is_palindromic(n): return str(n)==str(n)[::-1]
 
p=0
for i in range(999, 101, -2):
  for j in range(i, 101, -2):
    if i*j < p: break
    if is_palindromic(i*j): p = i*j
print "Answer to PE4 = ",p

Comments

Search took 3,072 iterations.
The largest palindrome for some other cases: 4 digits: 9999 × 9901 = 99000099, 5 digits: 99979 × 99681 = 9966006699, 6 digits: 999999 × 999001 = 999000000999, 7 digits: 9997647 × 9998017 = 99956644665999

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

What is the largest prime factor of the number 600851475143 ?

Analysis

The request to find the prime factors of a number comes up much too often. To satisfy these requests we wrote a small javascript program that can be accessed here: prime factor calculator. It can easily handle numbers up to 16 digits long in an instant.

The solution below is the essence of that script.

Solution

Runs < 1 second in Python.

def lpf(n):
  if n < 2: return False
  for p in (2,3):
    while n % p == 0: n /= p
    if n == 1: return p
 
  p = 5; fl = 1
  while (p*p<=n): 
    while n % p == 0: n /= p
    if n == 1: return p
    p+= 3-fl
    fl = -fl
  return n
print "Answer to PE3 = ",lpf(600851475143)

Comments

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, …

Find the sum of all the even-valued terms in the sequence which do not exceed four million.

Analysis

The approximate ratio between two consecutive terms in the Fibonacci sequence is the golden ratio (phi, or φ ≈ 1.618034). It can also be demonstrated that every 3rd term of the sequence is even. By combining this knowledge we can calculate the series of even Fibonacci numbers by multiplying the previous even term in the series by the cube of the golden ratio.

Solution

Runs < 1 second in Python.

limit = 4000000
phi_cubed = ((1+5**.5)/2)**3   #golden ratio cubed
 
f = 2
sum = 0
while f < limit:
  sum += f
  f = round(f*phi_cubed)
print "Answewr to PE2 - ",int(sum)

Comments

For a limit of 4,000,000 the loop iterates 10 times.

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.

Analysis

This is an ideal solution for the inclusion-exclusion principle. In general, we sum the numbers divisible by 3 or 5 for all numbers less than n, where n=1000 for this problem. Next, we would need to subtracxt those numbers divisible by 15 because they are counted twice.

This solution is very extensible instead of using brute force that requires loops.

Solution

Runs < 1 second in Python.

def sumn(n,d):
  n=int(n/d)
  return int(n*(n+1) / 2*d)
 
n = 999
print "Answer to PE1 = ",sumn(n,3) + sumn(n,5) - sumn(n,15)

Comments

The formula to sum the counting numbers from 1 to n is: n(n+1)/2. As for n=10; 10*11/2=55

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6^th prime is 13.

What is the 10001^st prime number?

Analysis

Having a sufficiently fast is_prime() funtion makes solving this problem quick and easy. Just enumerate the positive odd integers from 7 that follow the sequence 6m±1 until the n^th prime is found. For this problem n = 10001.

Solution

Runs < 1 second in Python.

from Euler import is_prime  
 
n = 10001
p = 5; f = 1
while n>3:   # prime numbers  2, 3 and 5 are accounted for
  p += 3-f
  f = -f
  if is_prime(p): n -= 1
print "Answer to PE7 = ", p

Comments

Click here for more information on the is_prime() function.
It took 34,913 iterations to solve this problem. Checking just odd numbers from 3 by 2′s takes 52,371 iterations.

A Pythagorean triplet is a set of three natural numbers, a ≤ b < c, for which,

For example, 3² + 4² = 9 + 16 = 25 = 5².

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

Analysis

This site provided a simple formula for generating Pythagorean triplets. Here’s the formula:
Suppose that m and n are two positive integers, with m < n.
Then n² – m², 2mn, and n² + m² is a Pythagorean triple.

Ok, let’s work this out:
(n² -m²) + 2mn + (n² + m²) = 1000
2n² + 2mn = 1000
2n(n+m) = 1000
n(n+m) = 500
for m=20, a
n² + mn = 500

So, to find the maximum n, such that m < n, we search downwards and return the first one found. An assumption for this method is that the gcd(a, b, c) = 1. This simply means that a, b and c need to be reduced to their lowest terms. For example, {6, 8, 10} is a Pythagorean triple, but can be reduced to {3, 4, 5}.

Solution

Runs < 1 second in Python.

def pe9(p):
  LL = int(p**.5)
  for i in range(1, LL, 2):
    for j in range(2, LL, 2):
        a = abs(j*j - i*i)
        b = 2*i*j;
        c = i*i + j*j
        if a+b+c == p:
           return a*b*c
print "Answer to PE9 = ",pe9(1000)

Comments