# Category Archives: Project Euler

## Longest Collatz sequence

The following iterative sequence is defined for the set of positive integers:

n n/2 (n is even)
n 3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13 40 20 10 5 16 8 4 2 1

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

Solution:

```
#include <iostream>

int main ()
{
int N           = 1000000;
int CHECKPOINT  = 800000;
long long nTemp = N;
int chainCount  = 1;
int check       = 0;
long long myNum = 0;

for (int i = N; i > CHECKPOINT; --i)
{
while (nTemp != 1)
{
if ((nTemp%2) == 0)
{
nTemp = nTemp/2;
}else
{
nTemp = (nTemp*3) + 1;
}
++chainCount;
}

--N;
nTemp = N;

if (check < chainCount)
{
check = chainCount;
myNum = nTemp;
}
chainCount = 1;
}

std::cout << "The number " << myNum << " has longest chain of " << check << std::endl;
}
```

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## Project Euler Problem#13 solution in C++

Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.

37107287533902102798797998220837590246510135740250
46376937677490009712648124896970078050417018260538
74324986199524741059474233309513058123726617309629
91942213363574161572522430563301811072406154908250
23067588207539346171171980310421047513778063246676
89261670696623633820136378418383684178734361726757
28112879812849979408065481931592621691275889832738
44274228917432520321923589422876796487670272189318
47451445736001306439091167216856844588711603153276
70386486105843025439939619828917593665686757934951
62176457141856560629502157223196586755079324193331
64906352462741904929101432445813822663347944758178
92575867718337217661963751590579239728245598838407
58203565325359399008402633568948830189458628227828
80181199384826282014278194139940567587151170094390
35398664372827112653829987240784473053190104293586
86515506006295864861532075273371959191420517255829
71693888707715466499115593487603532921714970056938
54370070576826684624621495650076471787294438377604
53282654108756828443191190634694037855217779295145
36123272525000296071075082563815656710885258350721
45876576172410976447339110607218265236877223636045
17423706905851860660448207621209813287860733969412
81142660418086830619328460811191061556940512689692
51934325451728388641918047049293215058642563049483
62467221648435076201727918039944693004732956340691
15732444386908125794514089057706229429197107928209
55037687525678773091862540744969844508330393682126
18336384825330154686196124348767681297534375946515
80386287592878490201521685554828717201219257766954
78182833757993103614740356856449095527097864797581
16726320100436897842553539920931837441497806860984
48403098129077791799088218795327364475675590848030
87086987551392711854517078544161852424320693150332
59959406895756536782107074926966537676326235447210
69793950679652694742597709739166693763042633987085
41052684708299085211399427365734116182760315001271
65378607361501080857009149939512557028198746004375
35829035317434717326932123578154982629742552737307
94953759765105305946966067683156574377167401875275
88902802571733229619176668713819931811048770190271
25267680276078003013678680992525463401061632866526
36270218540497705585629946580636237993140746255962
24074486908231174977792365466257246923322810917141
91430288197103288597806669760892938638285025333403
34413065578016127815921815005561868836468420090470
23053081172816430487623791969842487255036638784583
11487696932154902810424020138335124462181441773470
63783299490636259666498587618221225225512486764533
67720186971698544312419572409913959008952310058822
95548255300263520781532296796249481641953868218774
76085327132285723110424803456124867697064507995236
37774242535411291684276865538926205024910326572967
23701913275725675285653248258265463092207058596522
29798860272258331913126375147341994889534765745501
18495701454879288984856827726077713721403798879715
38298203783031473527721580348144513491373226651381
34829543829199918180278916522431027392251122869539
40957953066405232632538044100059654939159879593635
29746152185502371307642255121183693803580388584903
41698116222072977186158236678424689157993532961922
62467957194401269043877107275048102390895523597457
23189706772547915061505504953922979530901129967519
86188088225875314529584099251203829009407770775672
11306739708304724483816533873502340845647058077308
82959174767140363198008187129011875491310547126581
97623331044818386269515456334926366572897563400500
42846280183517070527831839425882145521227251250327
55121603546981200581762165212827652751691296897789
32238195734329339946437501907836945765883352399886
75506164965184775180738168837861091527357929701337
62177842752192623401942399639168044983993173312731
32924185707147349566916674687634660915035914677504
99518671430235219628894890102423325116913619626622
73267460800591547471830798392868535206946944540724
76841822524674417161514036427982273348055556214818
97142617910342598647204516893989422179826088076852
87783646182799346313767754307809363333018982642090
10848802521674670883215120185883543223812876952786
71329612474782464538636993009049310363619763878039
62184073572399794223406235393808339651327408011116
66627891981488087797941876876144230030984490851411
60661826293682836764744779239180335110989069790714
85786944089552990653640447425576083659976645795096
66024396409905389607120198219976047599490197230297
64913982680032973156037120041377903785566085089252
16730939319872750275468906903707539413042652315011
94809377245048795150954100921645863754710598436791
78639167021187492431995700641917969777599028300699
15368713711936614952811305876380278410754449733078
40789923115535562561142322423255033685442488917353
44889911501440648020369068063960672322193204149535
41503128880339536053299340368006977710650566631954
81234880673210146739058568557934581403627822703280
82616570773948327592232845941706525094512325230608
22918802058777319719839450180888072429661980811197
77158542502016545090413245809786882778948721859617
72107838435069186155435662884062257473692284509516
20849603980134001723930671666823555245252804609722
53503534226472524250874054075591789781264330331690

Solution:

```
#include <stdio.h>
#include <iostream>
#include "projectEulerProblem13.h" //data is stored in char data in this file.

int main ()
{
int x, y, columnSum, carry = 0;

for (x = 49; x >= 0; x--)
{
columnSum[x] = carry;
for (y = 0; y < 100; y++)
{
columnSum[x] += (int)data[y][x] - 48;
}

carry = columnSum[x] / 10;
}

std::cout << carry << std::endl;

for (x = 0; x < 50; x++)
{
std::cout << (columnSum[x] % 10) << std::endl;
}

return 0;
}

```

## Project Euler Problem#12 solution in C++

The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

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

Let us list the factors of the first seven triangle numbers:

1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28

We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

Solution:

```
#include <cstdio>
#include <cmath>

int numDivisors(int num)
{
int divisors = 1, exp;

//handle 2 and 3 as special case
exp = 0;
while (num % 2 == 0) {
num /= 2;
exp++;
}
divisors *= (exp+1);

exp = 0;
while (num % 3 == 0) {
num /= 3;
exp++;
}
divisors *= (exp+1);

//the other primes are 6k +/- 1
int i = 5;
int root = floor(sqrt(num));
while (i <= root) {
exp = 0;
while (num % i == 0) {
num /= i;
exp++;
}
divisors *= (exp+1);

if (i % 6 == 1)
i += 4;
else
i += 2;
root = floor(sqrt(num));
}

//a single largest prime factor
if (num != 1)
divisors *= 2;

return divisors;
}

int main(void)
{
int n = 0;
int divisors = 0;

while (divisors <= 500) {
//while (n <= 10) {
n++;

//since t = n*(n+1)/2 and gcd(n,n+1)=1
if (n % 2 == 0)
divisors = numDivisors(n/2) * numDivisors(n+1);
else
divisors = numDivisors(n) * numDivisors((n+1)/2);
}

printf("%d - %ld: %d\n", n, (long)n*(n+1)/2, divisors);
}
```

## Project Euler Problem#11 solution in C++

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?

Solution:

```#include<iostream>

const int num = 20;

void multiply(long long&, int [][num], int [], int, int, int ,int);

int main()
{

int array[num][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,  1, 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
};

long long maxNum = 1;
int bArray;
for (int i = 0; i < num; i++)
{
for (int j = 0; j < num; j++)
{
multiply(maxNum, array, bArray, i, j, 1, 0);
multiply(maxNum, array, bArray, i, j, 0, 1);
multiply(maxNum, array, bArray, i, j, 1, 1);
multiply(maxNum, array, bArray, i, j, -1, 1);
}
}

std::cout << maxNum << " = " << bArray << " * " << bArray << " * " << bArray << " * " << bArray << std::endl;

return 0;
}

void multiply(long long& maxNum, int array[][num], int bArray[], int x, int y, int dx, int dy)
{
long long tempNum = 1;
for (int i = 0; i < 4 && x < num && x >=0 && y < num && y >= 0; i++, x += dx, y += dy)
tempNum *= array[x][y];

if (maxNum < tempNum)
{
maxNum = tempNum;
bArray = array[x - dx][y - dy];
bArray = array[x - dx * 2][y - dy * 2];
bArray = array[x - dx * 3][y - dy * 3];
bArray = array[x - dx * 4][y - dy * 4];
}
}
```

## Project Euler Problem#10 solution in C++

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

Explanation:

Sieve of Eratosthenes: algorithm steps for primes below 121 (including optimization of starting from prime’s square).

In mathematics, the sieve of Eratosthenes (Greek: κόσκινον Ἐρατοσθένους), one of a number of prime number sieves, is a simple, ancient algorithm for finding all prime numbers up to any given limit. It does so by iteratively marking as composite (i.e. not prime) the multiples of each prime, starting with the multiples of 2.

The multiples of a given prime are generated starting from that prime, as a sequence of numbers with the same difference, equal to that prime, between consecutive numbers. This is the sieve’s key distinction from using trial division to sequentially test each candidate number for divisibility by each prime.

The sieve of Eratosthenes is one of the most efficient ways to find all of the smaller primes (below 10 million or so). It is named after Eratosthenes of Cyrene, a Greek mathematician; although none of his works have survived, the sieve was described and attributed to Eratosthenes in the Introduction to Arithmetic by Nicomachus. (above explanation taken from Wikipedia)

C++ solution:

//solution is written in a flexible manner that you could compute sum of n prime numbers.

#include <iostream>

using namespace std;

int find_prime (long *numArray, int maxNum)
{
long factor = 2; // we will make 2 as the starting point.
numArray = 0; // rule out 1 from our logic to avoid incorrect results.

// loop condition will check, if we are in our maximum number limit.
//maxNum is the number till there we can find the prime numbers.
while ((factor * factor) < maxNum)
{
for (int i = 2; i < maxNum; i++) // we start our itration from number 2 not from 0 or 1 to get correct results.
{
if (numArray[i] != 0) //if a number on current array index is not zero, it is a prime or we havne’t yet checked it.
{
// we are putting zeros on all multiples of prime numbers, one by one.
if (numArray[i] != factor && (numArray[i]% factor) == 0)
{
numArray[i] = 0;
}
}
}
++factor;
}// after the loop, array will have zeros on all non prime locations and prime numbers.
}

int main ()
{
int prime_count = 0;
int64_t sum = 0;
int maxNum = 0;
int i = 0;

cout << “enter max number for array: “; // you need to test some upper limit values.
cin >> maxNum;

long *myArray = new long [maxNum];

//we fill up the array till the number we want to find the smallest positive number that is evenly divisible.
for (int i = 0; i < maxNum; i++)
{
myArray[i]= i;
}

// we will get prime numbers till the maxNum by calling below funtions.
find_prime(myArray, maxNum);

for (int j = 0; j< maxNum; j++)
{
sum += myArray[j];
}

cout << “Sum of first ” << maxNum <<  ” prime numbers : ” << sum << endl;

}

## Project Euler Problem# 8 solution in C++

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

Solution:

#include <iostream>
#include <cstdlib>

using namespace std;

int main()
{
int maxNum =1;
char a ;
string test ;
int result =1;
int checker = 1;

string numString  = (“7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450”);

for (int i = 0; i < 1000; i++)
{
if (numString[i] == ‘9’)
{
for (int j = 0; j < 5; j++)
{
a = numString[i+j];
maxNum *= atoi(&a);
}
if (result < maxNum)
{
result = maxNum;
}else
{
cout << result << endl;
}
maxNum = 1;
}
}
}

## Project Euler Problem#7 solution in C++ (Brute Force)

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?

Brute Force Solution:

#include <iostream>

using namespace std;

int find_prime (long *numArray, int maxNum)
{
long factor = 2; // we will make 2 as the starting point.
numArray = 0; // rule out 1 from our logic to avoid incorrect results.

// loop condition will check, if we are in our maximum number limit.
//maxNum is the number till there we can find the prime numbers.
while ((factor * factor) < maxNum)
{
for (int i = 2; i < maxNum; i++) // we start our itration from number 2 not from 0 or 1 to get correct results.
{
if (numArray[i] != 0) //if a number on current array index is not zero, it is a prime or we havne’t yet checked it.
{
// we are putting zeros on all multiples of prime numbers, one by one.
if (numArray[i] != factor && (numArray[i]% factor) == 0)
{
numArray[i] = 0;
}
}
}
++factor;
}// after the loop, array will have zeros on all non prime locations and prime numbers.
}

int main ()
{
int prime_count = 0;
int nth_prime = 0;
int maxNum = 0;
int i = 0;

cout << “enter max number for array: “; // you need to test some upper limit values.
cin >> maxNum;
cout << endl << “enter nth prime number you are searching for: “;
cin >> nth_prime;

long *myArray = new long [maxNum];

//we fill up the array till the number we want to find the smallest positive number that is evenly divisible.
for (int i = 0; i < maxNum; i++)
{
myArray[i]= i;
}

// we will get prime numbers till the maxNum by calling below funtions.
find_prime(myArray, maxNum);

for (; i < maxNum; i++)
{
if (myArray[i] != 0)
{
prime_count++;
}
if (prime_count == nth_prime){break;}
}

cout << “the nth prime number is: ” << “value of prime count is: “<< prime_count << ” – ” << myArray[i] << endl;

}

## Project Euler Problem# 6 solution in C++

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.

Efficient solution without using any loops. So it will be constant time method:

#include<iostream>

using namespace std;

int main ()
{
int maxNum = 0;
int sum_of_square = 0;
int square_of_sum = 0;

cout << “Enter Maximum number: “;
cin >> maxNum;

sum_of_square = ((maxNum*(maxNum +1))*((2*maxNum)+1))/6;

square_of_sum *= square_of_sum = maxNum* (maxNum+1)/2;

cout << “sum of squares till ” << maxNum << ” is: ” << sum_of_square << endl;
cout << “squares of sum till ” << maxNum << ” is: ” << square_of_sum << endl;
cout << “difference between the sum of the squares the square of the sum: ” << square_of_sum – sum_of_square  << endl;
}

## Project Euler Problem# 5 solution in C++

/**
Project Euler problem# 5:
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?

Below is an efficient solution of above problem using C++. Time complexity in the random access machine model is O(n log log n)
operations, a direct consequence of the fact that the prime harmonic series asymptotically approaches log log n.

*/

#include <iostream>

using namespace std;

//Finding Prime Numbers using the sieve of Eratosthenes, one of a number of prime number sieves,
//is a simple, ancient algorithm for finding all prime numbers up to any given limit.
// it is not an trial and check algorithm but find prime numbers by jumping out.
//The sieve of Eratosthenes is one of the most efficient ways to find all of the smaller primes (below 10 million or so)
int find_prime (int *numArray, int maxNum)
{
int factor = 2; // we will make 2 as the starting point.
numArray = 0; // rule out 1 from our logic to avoid incorrect results.

// loop condition will check, if we are in our maximum number limit.
//maxNum is the number till there we can find the prime numbers.
while ((factor * factor) < maxNum)
{
for (int i = 2; i < maxNum; i++) // we start our itration from number 2 not from 0 or 1 to get correct results.
{
if (numArray[i] != 0) //if a number on current array index is not zero, it is a prime or we haven’t yet checked it.
{
// we are putting zeros on all multiples of prime numbers, one by one.
if (numArray[i] != factor && (numArray[i]% factor) == 0)
{
numArray[i] = 0;
}
}
}
++factor;
}// after the loop, array will have zeros on all non prime locations and prime numbers.
}

int main ()
{
int maxNum = 0;
int result = 1;
int num = 0;

cout << “enter max number: “;
cin >> maxNum;

int *myArray = new int [maxNum];

//we fill up the array till the number we want to find the smallest positive number that is evenly divisible.
for (int i = 0; i < maxNum; i++)
{
myArray[i]= i;
}

// we will get prime numbers till the maxNum by calling below function.
find_prime(myArray, maxNum);

// returning array will have prime numbers and zeros on non prime locations.
// we start our loop from 2 as first two locations are not use full so we reduce the itration of the loop.
for (int i = 2; i < maxNum; i ++)
{
// we only do our calculation on prime numbers.
if ((myArray [i]) != 0)
{
num = myArray[i];
// we take multiples of each prime number till the maxNum. meaning we raise each prime to power of a number.
// which result will remain till maxNum.
while ((num * (myArray[i])) < maxNum)
{
num = num * (myArray[i]);
}
result = num * result;
}
}// after the loop result will have smallest positive number that is evenly divisible by all of the numbers from 1 to maxNum.
cout << “smallest positive number that is evenly divisible by all of the numbers from 1 to ” << maxNum
<< ” is: ” << result <<endl;
delete []myArray;
}

## Project Euler Problem #2 Solution in Java

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.

/**
* @author Khuram Ali
*/
public class Problem2
{
public static void main(String[] args)
{
int term_1 = 1;
int term_2 = 2;
int term_3 = 0;
int sum = 0;
long MaxTerm = 4000000;

while (term_1 <= MaxTerm)
{
System.out.println (term_1);
if ((term_1 %2) == 0){sum += term_1;}

term_3 = term_1 + term_2;
term_1 = term_2;
term_2 = term_3;
}
System.out.println (“the sum of even value numbers is: “);
System.out.println (sum);
}
}