Posts Tagged ‘Mathematics’

Project Euler: Problem 8

Monday, August 9th, 2010

Problem 8 in Project Euler is this

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

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Project Euler: Problem 7

Sunday, August 8th, 2010

This is problem 7 on Project Euler:

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?

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Project Euler: Problem 6

Saturday, August 7th, 2010

This is problem 6 on Project Euler:

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

1^(2) + 2^(2) + … + 10^(2) = 385

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

(1 + 2 + … + 10)^(2) = 55^(2) = 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.

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Three Circles in a Square

Friday, August 6th, 2010

Taking a break from Project Euler I have come across this problem recently

It is obvious that we can fit four circles of diameter 1 unit in a square of side 2 units without overlapping. What is the smallest square into which we can fit 3 circles of diameter 1 unit?

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