The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another.

There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.

What 12-digit number do you form by concatenating the three terms in this sequence?

Analysis

An obvious solution. Start with the known solution plus 2 (1487+2) and find the next set checking only odd numbers.

Solution

Runs < 1 second in Python.

from Euler import is_prime, is_perm
 
n = 1489  # must be odd
while True:
  b, c = n+3330, n+6660
  if is_prime(n) and is_prime(b) and is_prime(c) \
    and is_perm(n,b) and is_perm(b,c): break
  n += 2
print "Answer to PE49 = ", str(n)+str(b)+str(c)

Comments

• More information on the Euler module can be found on the tools page.
• It wasn’t clear, perhaps intentionally, that the terms in the sequence would increase by the same amount (3330) as the example.

The first two consecutive numbers to have two distinct prime factors are:

14 = 2 × 7
15 = 3 × 5

The first three consecutive numbers to have three distinct prime factors are:

644 = 2² × 7 × 23
645 = 3 × 5 × 43
646 = 2 × 17 × 19.

Find the first four consecutive integers to have four distinct primes factors. What is the first of these numbers?

Analysis

Simple approach that took only a few minutes to write. Start with the first number that has 4 distinct prime factors. Continue to increment from that number until 4 consecutive numbers have four distinct prime factors.

Solution

Runs < 5 seconds in Python.

from Euler import factor
 
c = 1
n = 2 * 3 * 5 * 7
while c != 4:
  n += 1
  if len(factor(n)) == 4: c += 1
  else: c = 0
print "Answer to PE47 = ", n-3

Comments

More information on the Euler module can be found on the tools page.

It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.

9 = 7 + 2×1²
15 = 7 + 2×2²
21 = 3 + 2×3²
25 = 7 + 2×3²
27 = 19 + 2×2²
33 = 31 + 2×1²

It turns out that the conjecture was false.

What is the smallest odd composite that cannot be written as the sum of a prime and twice a square

Analysis

Generate primes as needed and check the conjecture until a prime isn’t found.

Solution

Runs < 1 second in Python.

n = 5
f = 1
primes = set()
 
while (1):
  if all( n % p for p in primes ):
    primes.add(n)
  else:
    if not any( (n-2*i*i) in primes for i in xrange(1, n) ):
      break
  n += 3-f
  f = -f
print "Answer to PE46 = ", n

Comments

An irrational decimal fraction is created by concatenating the positive integers:

0.123456789101112131415161718192021…

It can be seen that the 12^th digit of the fractional part is 1.

If d_n represents the n^th digit of the fractional part, find the value of the following expression.

d₁ × d₁₀ × d₁₀₀ × d₁₀₀₀ × d₁₀₀₀₀ × d₁₀₀₀₀₀ × d_1000000

Analysis

This is known as Champernowne’s constant (OEIS A033307).

We concatenate the counting numbers to a string from 1 until the length reaches at least 1 million digits, then multiply the digit at the appropriate positions together. We know from the problem description that the first two terms d₁ & d₁₀ will evaluate to 1 and, therefore, serve no consequence on the product.

Solution

Runs < 1 second in Perl.

my $s = '.';
my $n = 1;
 
$s.=$n++ while(length($s) <= 1_000_000);
 
print "Answer to PE40 = ",
substr($s,1e2,1) * substr($s,1e3,1) * substr($s,1e4,1) * substr($s,1e5,1) * substr($s,1e6,1);

Comments

• We initialize $s with a character to start the string index at 1 instead of 0.
• Mathematica: Times @@ Flatten[IntegerDigits@Range@2*^5][[10^Range@6]]

The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.

Let d₁ be the 1^st digit, d₂ be the 2^nd digit, and so on. In this way, we note the following:

• d₂d₃d₄=406 is divisible by 2
• d₃d₄d₅=063 is divisible by 3
• d₄d₅d₆=635 is divisible by 5
• d₅d₆d₇=357 is divisible by 7
• d₆d₇d₈=572 is divisible by 11
• d₇d₈d₉=728 is divisible by 13
• d₈d₉d₁₀=289 is divisible by 17

Find the sum of all 0 to 9 pandigital numbers with this property.

Analysis

There are only 9 x 9! (3,265,920) possible 0-9 10-digit pandigital numbers. Simply generate the possibilities and check each 3-digit section to be divisible by the appropriate prime number. But even on a fast computer this method broke the one minute limitation.

Clearly something had to be done to reduce the number of permutations and upon closer inspection some rules were discovered. Keep in mind we are dealing with 3 digit numbers that must have unique digits in order to qualify as part of a 0-9 pandigital 10-digit number.

Rule 0: Can’t start with {0}.

Rule 1: Since d₄ has to be divisible by 2 then d₄ has to be from the set {0, 2, 4, 6, 8}.

Rule 2: Since d₄d₅d₆ has to be divisible by 5 then d₆ has to be from the set {0, 5}.

Rule 3: d₆d₇d₈ has to be divisible by 11 and, according to Rule 2, has to start with a {0, 5}. Well, it can’t start with 0 as that would yield invalid numbers such as {011, 022, … 099}, so it has to start with 5 and have only unique digits. The set is
{506, 517, 528, 539, 561, 572, 583, 594}.

New Rule 2: Amend Rule 2 to d₆ = {5}.

Rule 5: Since d₇d₈d₉ must be divisible by 13, begin with {06, 17, 28, 39, 61, 72, 83, 94}, contain no 5s and no repeated digits our set becomes {286, 390, 728, 832}.

Rule 6: Since d₈d₉d₁₀ must be divisible by 17, begin with {28, 32, 86, 90}, contain no 5s and no repeated digits our set becomes {289, 867, 901}.

Let’s review what we have and what combinations are possible for d₆ through d₁₀:
d₆ = {5}
d₇d₈d₉ = {286, 390, 728}
d₈d₉d₁₀ = {289, 867, 901}

d₆d₇d₈d₉d₁₀ = {52867, 53901, 57289}

Rule 7: d₅d₆d₇ must be divisible by 7 and end with {52, 53, 57} yields {357, 952}

Rule 1: Amend Rule 1 to d₄ = {0, 4, 6}.

Our solution set ends with {….357289 or ….952867}

You could finish the problem furthering this technique, but we reduced it enough to have the computer solve the rest.

Solution

Runs < 1 second in Python.

def fact(n):
  return reduce(lambda x,y:x*y,range(1,n+1),1)
 
def perm(n, s):
  if len(s)==1: return s
  q, r = divmod(n, fact(len(s)-1))
  return s[q] + perm(r, s[:q] + s[q+1:])
 
def check(n):
  s = str(n)
  return int(s[1:4]) % 2 == 0 and int(s[2:5]) % 3 == 0
 
s = 0
for i in range(0,18):
  a = int( perm(i,'4310')+'952867')
  if check(a): s += a
  a = int( perm(i,'6410')+'357289')
  if check(a): s += a
print "Answer to PE43 = ", s

Comments

The n^th term of the sequence of triangle numbers is given by, t_n = ½n(n+1); so the first ten triangle numbers are:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, …

By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is 19 + 11 + 25 = 55 = t₁₀. If the word value is a triangle number then we shall call the word a triangle word.

Using words.txt (right click and ‘Save Link/Target As…’), a 16K text file containing nearly two-thousand common English words, how many are triangle words?

Analysis

Remember Problems 44 & 45? The same logic applies except for triangle numbers. Whenever the equation ( √(1+8n) – 1 ) / 2 evaluates to an integer the number, n, is a triangle number.

Remember Problem 22? Well, this is the same core solution.

Solution

Runs < 1 second in Perl.

$c = 0;
@h{'A' .. 'Z'} = (1 .. 26);
open (IN,"<words.txt") or die "Can't open file: $!";
@words = map { /"(w+)"/  } split /,/,<IN>;
 
foreach ( grep { is_tri(score($_)) } @words) {$c++}
 
print "Answer to PE42 = $c";
 
sub is_tri {my $n=shift; my $t=(sqrt(1+8*$n)-1)/2; return int($t) == $t }
sub score {my $score=0; $score+=$_ for @h{map uc, split //,shift}; return $score };

Comments

Even though we used Perl, we still kept it readable, mostly.

Pentagonal numbers are generated by the formula, P_n=n(3n−1)/2. The first ten pentagonal numbers are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, …

It can be seen that P₄ + P₇ = 22 + 70 = 92 = P₈. However, their difference, 70 − 22 = 48, is not pentagonal.

Find the pair of pentagonal numbers, P_j and P_k, for which their sum and difference is pentagonal and D = |P_k − P_j| is minimised; what is the value of D?

Analysis

How many pentagonal numbers are required to search through to solve the problem? We wrote the following program and started with 10,000 pentagonal numbers and, after finding the correct answer, adjusted the requirement down to 2,400. When the problem asks for D = |P_k − P_j| to be minimized it simply means the smallest pair which, by our method, is the first occurrence found.

A set (hashable) is used to both hold the pentagonal numbers and to check if a number is pentagonal. All that’s left is to permute the pairs and check that the difference and sum are pentagonal numbers.
Runs < 2 seconds in Python.

pent = set( n * (3*n - 1) / 2 for n in range(2,2400) ) 
 
def pe44():
  for pj in pent: 
    for pk in pent: 
      if pj - pk in pent and pj + pk in pent: 
        return pj - pk
 
print "Answer to PE44 = ",pe44()

So, a more challenging program might be to solve the problem without having to guess at how many pentagonal numbers are required. That would require a function to determine if a number is pentagonal. (When the equation (√(1 + 24 · n) + 1) / 6 evaluates to an integer, n, is a pentagonal number.)

It turned out to be an interesting discovery and took longer to find an answer. But, as luck would have it, was very helpful for solving problem 45.

Solution

Runs < 6 seconds in Python.

from math import sqrt
 
def is_pentagonal(n):
  p = ( sqrt(1 + 24*n) + 1 ) / 6
  return p==int(p)
 
def pe44():
  i=0;
  pent = [1]
  while True:
    i+=1
    pent.append( i * (3*i - 1) / 2 )
    for j in range(2, len(pent)-1):
      if is_pentagonal(pent[i]-pent[j]) and is_pentagonal(pent[i]+pent[j]):
        return pent[i]-pent[j]
 
print "Answer to PE44 = ",pe44()

Comments

See Problem 45.

Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:

It can be verified that T₂₈₅ = P₁₆₅ = H₁₄₃ = 40755.

Find the next triangle number that is also pentagonal and hexagonal.

Analysis

Note that Hexagonal numbers are a subset of Triangle numbers so we only determine the first occurrence of H_i = P_j to find our answer. We can safely start our search from 166 as eluded to in the problem’s description.

From our experience with problem 44 we were able to make our search trivial without the use of arrays.

Solution

Runs < 1 second in Python.

from math import sqrt
 
def is_pentagonal(n):
  p = ( sqrt(1 + 24*n) + 1 ) / 6
  return p==int(p)
 
n = 165 
while True:
  n += 1 
  hex = n * (2 * n-1)
  if is_pentagonal(hex): break 
 
print "Answer to PE45 = ",hex

Comments

See Problem 44.
The next number in the series is 57722156241751, calculated in about 15 seconds.

We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.

What is the largest n-digit pandigital prime that exists?

Analysis

What is the value of n? We know it’s between 4 and 9 but that leaves a huge set of prime number candidates to search through. So let’s eliminate some values of n by using the rule that any integer is divisible by 3 or 9 whose sum of digits is divisible by 3 or 9; therefore composite and not prime.

9+8+7+6+5+4+3+2+1 = 45,
8+7+6+5+4+3+2+1 = 36,
6+5+4+3+2+1 = 21, and
5+4+3+2+1 = 15,
all of which are divisible by 3 and therefore could not yield a 1 to {5, 6, 8, 9} pandigital prime. So that leaves 4 and 7 digit prime numbers less than 4321 and 7654321 respectively. Since we want the largest pandigital prime we’ll check the 7 digit group first.

Solution

Runs < 1 second in Python.

from Euler import is_prime, is_pandigital
 
for n in xrange(7654321, 1234567, -2):
  if is_prime(n) and is_pandigital(n, 7):
    print "Answer to PE41 = ", n
    break

Comments

• BTW, the largest 4 digit pandigital prime is 4231.
• More information on the Euler module can be found on the tools page.

The series, 1¹ + 2² + 3³ + … + 10¹⁰ = 10405071317.

Find the last ten digits of the series, 1¹ + 2² + 3³ + … + 1000¹⁰⁰⁰.

Analysis

Without Python’s built-in pow() function and native large integer support this may have been more of a challenge. In fact, before revisiting this problem we solved it in Perl with the following code.

$max_n = 1000;
$m = 1e10;
$sum = 0;
 
for $n (1..$max_n) { 
	$nx = 1; 
	$nx = ($nx * $n) % $m  for (1..$n); 
	$sum = ($sum + $nx) % $m; 
}
 
print "PE 48 = $sum";

Solution

Runs < 1 second in Python.

def PE48(limit, m):
    sum = 0
    while limit:
        sum += pow(limit, limit, 10**m)
        limit -= 1
    return sum % 10**m
 
print "Answer to PE48 = ",PE48(1000, 10)

Comments

Note: (a*a*a) % n = ( (a%n) * (a%n) * (a%n) ) % n

• See also problem 97 & problem 188.
• Don’t forget that leading zeros are not printed.
• The last iteration, 1000, would not affect the last 10 digits of the sum, but was included to adhere to the problem description.
• The full answer is 3001 digits.