Someone told me we need more real programmers around, who actually understand what’s going on under the hood, and not the type who have no idea what they are building. The IT world doesn’t need more programmers who write a code patched around from different ‘Googleing missions’, only to break down when a website receives more traffic, and to put the business owners in the situation when they have to call a really skilled person to fix the bugs.

I don’t want to be part of the problem. I don’t want to just scratch the surface, I want to know it all. The way I went, so far, about teaching myself to code made me the painter, in the whole process of building a house. Yes, you do get the applause at the end of the job, but you have no idea about how to build the foundation. Actually, your work would be nothing without the guy who does the foundation.

Looking for a way to train my brain to understand how a computer works behinds the scene, I had to go back to the basics. Yes, no more attempts to build fancy website, only boring numbers until I get so good at it, that I can do it with my eyes closed.

That being said, I took the** Project Euler** Challenge and I’ve decided to share my Knowledge with everyone, while I complete a few problems.

Back to the basics, using Javascript to solve the following:

**Problem 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.

**var sum = 0;**

** for (i=0; i<1000; i++) {**

** if (i%3 === 0 || i%5 === 0) {**

** sum += i;**

** }**

** }**

** console.log(sum);**

**Correct Answer: 233168**

**Problem 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.

**var sum = 0;**

** var x = 0;**

** var y = 1;**

**while (y<4000000) {**

** if (y%2 === 0) {**

** sum = sum + y;**

** }**

**z = x + y;**

** x = y;**

** y = z;**

** }**

** console.log(sum);**

**Correct Answer: 4613732**

**Problem 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 ?

**function isaprime(p) {**

** var k = p / 2;**

** for (i=2; i<k; i++) {**

** if (p % i === 0) {**

** return false;**

** }**

** }**

** return true;**

**}**

**n = 13195;**

**for (i=3; i<=n; i++) {**

** if (n % i === 0) {**

** if (isaprime(i)) {**

** console.log(i + ” ” + n);**

** }**

** if ( n != i ) {**

** n = n / i;**

** i = 3;**

** }**

** }**

**}**

**Correct Answer: 6857**

**Problem 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.

*function reverse (n) {*

* var r = ‘ ‘;*

* while (n.length>0) {*

* r += n.substring(n.length-1,n.length);*

* n = n.substring(0, n.length-1);*

* }*

* return r;*

* }*

*function ispalindrome(n) {*

* return reverse(n) == n;*

* }*

*m = 0;*

*for (i=100; i<=999; i++) {*

* for (j=100; j<=999; j++) {*

* n = i * j;*

* if (ispalindrome(n.toString())) {*

* if ( n > m ) {*

* m = n;*

* }*

* }*

* }*

* }*

* console.log(m);*

**Correct Answer: 906609**

**Problem 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?

*var i = 20;*

* while ( i % 2 !== 0 || i % 3 !== 0 || i % 4 !== 0 || i % 5 !== 0 || i % 6 !== 0 || i % 7 !== 0 || i % 8 !== 0 || i % 9 !== 0 || i % 10 !== 0 || i % 11 !== 0 || i % 12 !== 0 || i % 13 !== 0 || i % 14 !== 0 || i % 15 !== 0 || i % 16 !== 0 || i % 17 !== 0 || i % 18 !== 0 || i % 19 !== 0 || i % 20 !== 0 ) {*

* i += 20;*

* }*

* console.log(i);*

**Correct Answer: 232792560**

**Problem 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.

*function sumsq(n){*

* sum = 0;*

* for (i=1; i<n; i++) {*

* sum += Math.pow(i,2);*

* } return sum;*

*}*

*function sqsum(n) {*

* sum = 0;*

* for (i=1; i<n; i++) {*

* sum += Math.pow(sum,2);*

* } *

* }*

*console.log(sumsq(100)-sqsum(100));*

**Correct Answer: 25164150**

**Problem 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?

*function isaprime(p) {*

* if ( n<2 ) {*

* return false;*

* }*

* var n = Math.sqrt(p);*

* for (i=2; i<=n; i++) {*

* if (p % i === 0) {*

* return false;*

* }*

* }*

* return true;*

*}*

*var a = 0;*

*var primes = 0;*

*while (primes <= 10001) {*

* a++;*

* if (isaprime(a)) {*

* primes++;*

* }*

* }*

* console.log(a);*

**Correct Answer: 104743**

**Problem 8 – Largest product in a series**

The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.

73167176531330624919225119674426574742355349194934

96983520312774506326239578318016984801869478851843

85861560789112949495459501737958331952853208805511

12540698747158523863050715693290963295227443043557

66896648950445244523161731856403098711121722383113

62229893423380308135336276614282806444486645238749

30358907296290491560440772390713810515859307960866

70172427121883998797908792274921901699720888093776

65727333001053367881220235421809751254540594752243

52584907711670556013604839586446706324415722155397

53697817977846174064955149290862569321978468622482

83972241375657056057490261407972968652414535100474

82166370484403199890008895243450658541227588666881

16427171479924442928230863465674813919123162824586

17866458359124566529476545682848912883142607690042

24219022671055626321111109370544217506941658960408

07198403850962455444362981230987879927244284909188

84580156166097919133875499200524063689912560717606

05886116467109405077541002256983155200055935729725

71636269561882670428252483600823257530420752963450

Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

*var n = “7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450”;*

*m = 0;*

*for (j=0; j<=n.length – 13; j++) {*

* var p = n[j] * n[j+1] * n[j+2] * n[j+3] * n[j+4] * n[j+5] * n[j+6] * n[j+7] * n[j+8] * n[j+9] * n[j+10] * n[j+11] * n[j+12];*

* if ( p>m ) {*

* m = p;*

* }*

* }*

*console.log(Math.max(m));*

**Correct Answer: 23514624000**

**Problem 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.

*var a = 0;*

*var b = 0;*

*var c = 0;*

*var sum = 1000;*

*for (a=1; a<1000; a++) {*

* for (b=a+1; b<1000; b++) {*

* c = 1000 – a – b;*

* if (a*a + b*b == c*c) {*

* console.log(a, b , c);*

* console.log(a*b*c);*

* }*

* }*

*}*

**Correct Answer: 31875000**

**Problem 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.

*function isaprime(p) {*

* if ( n<2 ) {*

* return false;*

* }*

* var n = Math.sqrt(p);*

* for (i=2; i<=n; i++) {*

* if (p % i === 0) {*

* return false;*

* }*

* }*

* return true;*

*}*

*var sum = 0;*

*var x = 2000000;*

*for (j=2; j<x; j++) {*

* if (isaprime(j) == true) {*

* sum = sum + j;*

* }*

*}*

*console.log(sum);*

**Correct Answer: 142913828922**

**Problem 11 – Largest product in a grid**

In the 20×20 grid below, four numbers along a diagonal line have been marked in red.

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08

49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00

81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65

52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91

22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80

24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50

32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70

67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21

24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72

21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95

78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92

16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57

86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58

19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40

04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66

88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69

04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36

20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16

20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54

01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

The product of these numbers is 26 × 63 × 78 × 14 = 1788696.

What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?

*var num = [*

*[8,2,22,97,38,15,0,40,0,75,4,5,7,78,52,12,50,77,91,8],*

*[49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,4,56,62,0],*

*[81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,3,49,13,36,65],*

*[52,70,95,23,4,60,11,42,69,24,68,56,1,32,56,71,37,2,36,91],*

*[22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80],*

*[24,47,32,60,99,3,45,2,44,75,33,53,78,36,84,20,35,17,12,50],*

*[32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70],*

*[67,26,20,68,2,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21],*

*[24,55,58,5,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72],*

*[21,36,23,9,75,0,76,44,20,45,35,14,0,61,33,97,34,31,33,95],*

*[78,17,53,28,22,75,31,67,15,94,3,80,4,62,16,14,9,53,56,92],*

*[16,39,5,42,96,35,31,47,55,58,88,24,0,17,54,24,36,29,85,57],*

*[86,56,0,48,35,71,89,7,5,44,44,37,44,60,21,58,51,54,17,58],*

*[19,80,81,68,5,94,47,69,28,73,92,13,86,52,17,77,4,89,55,40],*

*[4,52,8,83,97,35,99,16,7,97,57,32,16,26,26,79,33,27,98,66],*

*[88,36,68,87,57,62,20,72,3,46,33,67,46,55,12,32,63,93,53,69],*

*[4,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36],*

*[20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,4,36,16],*

*[20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57,5,54],*

*[1,70,54,71,83,51,54,69,16,92,33,48,61,43,52,1,89,19,67,48]*

*];*

*var m = 0;*

*for (i = 0; i < num.length; i++) {*

* for (j = 0; j < num[i].length – 3; j++){*

* m = Math.max(num[i][j] * num[i][j+1] * num[i][j+2] * num[i][j+3],m);*

* }*

*}*

*for (i = 0; i < num.length – 3; i++) {*

* for (j = 0; j < num[i].length; j++){*

* m = Math.max(num[i][j] * num[i+1][j] * num[i+2][j] * num[i+3][j],m);*

* }*

* for (j = 0; j < num[i].length – 3; j++){*

* m = Math.max( num[i][j] * num[i+1][j+1] * num[i+2][j+2] * num[i+3][j+3],m);*

* }*

* for (j = 3; j < num.length; j++) {*

* m = Math.max(num[i][j] * num[i+1][j-1] * num[i+2][j-2] * num[i+3][j-3],m);*

* }*

*}*

*console.log(m);*

**Correct Answer: 70600674**

If you enjoyed this post, you should** Follow Me On Twitter** and **Like My Page on Facebook** for more updates.