I have been struggling with a couple of Euler Problems the past days. Not so much finding the solution as finding a couple of bugs in my code…
By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
3
7 4
2 4 6
8 5 9 3
Find the maximum total from top to bottom of the given triangle. Continue reading →
After my success on getting unittest to work on the iPhone, I thought I should try it out some more. It’s been a while since last time I solved an Euler Problem, so maybe this would be a good excuse to solve another one? The last problems involved a lot of math, so this time I found something completely different:
If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?
Well, lets see, then…
Yet another problem involving primes. Guess I have to make a better prime generator soon…
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
On our way to Lofoten last weekend, my wife and I had a small brainstorming on problem no. 9:
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 couple of days ago, I solved eulers problem no. 8. I finally found some time to blog about my solution.
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
It want be easy to keep up with my brother on this. Well, here’s my take on Project Eulers problem no 7:
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?
Inspired by my brother over at geekality.net, I thought I should do an attempt on the various problems presented at projecteuler.net/. Since he already solved the first five, I jumped right in at problem 6:
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.
