## Problem: Easy sum

*Difficulty: Easy

Problem:

Little Kevin had never heard the word ‘Infinitum’. So he asked his mentor to explain the word to him. His mentor knew that ‘Infinitum’ is a very large number. To show him how big Infinitum can be, his mentor gave him a challenge: to sum the numbers from 1 up to N. The sum started to get really large and was out of long long int range. And so the lesson was clear.

Now his mentor introduced him to the concept of mod and asked him to retain only the remainder instead of the big number. And then, he gave him a formula to compute:

Input Format
The first line contains T, the number of test cases.
T lines follow, each containing 2 space separated integers N m

Output Format
Print the result on new line corresponding to each test case.

Constraint
1 ≤ T ≤ 1000
1 ≤ N ≤ 109
1 ≤ m ≤ 109

Sample Input

3
10 5
10 3
5 5


Sample Output

20
10
10


Explanation
Case 1: N = 10 m = 5,
1%5 + 2%5 + 3%5 + 4%5 + 5%5 + 6%5 + 7%5 + 8%5 + 9%5 + 10%5 = 20.
Similar explanation follows for Case 2 and 3.

How to solve:

Base on the formula to calculate sum of first nth numbers (1+2+…+nth), and the property of mod. I came up with the formula below

init:

n <- N;

m <-m;

k = (floor)n/m;

m -= 1;

remainder = n%m;

if remainder == 0

result will be  ((m*(m+1))/2)*k

otherwise

result will be  ((m*(m+1))/2)*k + ((remainder*(remainder+1)/2))

As I said above, n*(n+1) / 2 is the formula to calculate the first n numbers (1,2,3,…n)

I will take two special cases that are the only two cases in this problem you need to pay attention.

Case 1: remainder == 0

n = 10, m = 5 (n mod 10 == 0)

Sum of first 10 number is :

1+2+4+5+6+7+8+9+10

Sum of first 10 number mod 5 is

1%5 + 2%5 + 3%5 + 4%5 + 5%5 + 6%5 + 7%5 + 8%5 + 9%5 + 10%5

As you may notice, 5%5 and 10%5 bot produce 0

and

1%5 + 2%5 + 3%5 + 4%5 = 6%5+ 7%5 + 8%5 + 9%5

It mean we can split the original expression into 2 equal part (n/m = 10/5 = 2)

And we just need to calculate sum of first (m-1) number, in this case is 5-1=4,

multiply that with k = n/m = 10/2 = 2.

Case 2: remainder != 0

We will use the value of case 1 and add it with the value produced by the formula we will discover here.

n = 10, m = 3

k = n/m = 10/3 (floor) = 3

remainder = 10 mod 3 = 1;

1%3 + 2%3 + 3%3 + 4%3 + 5%3 + 6%3 + 7%3 + 8%3 + 9%3 + 10%3

3%3 = 6%3 = 9%3 = 0

1%3 + 2%3 = 4%3 + 5%3 = 7%3 + 8%3  (3 parts = 10/3(floor) = 3)

10%3 = 1 (1 here is not the remainder of 10 mod 3, 1 is the sum of first 1 numbers)

We just need to calculate the sum of first (m-1), here is 3-1, numbers. And multiply that with 3 (as with Case 1) . About the remainder, 10%3 = 1%3 = sum of first 1 numbers

Another example:

n = 10, m  = 4

k = n / m = 2;

remainder = 2;

1%4+2%4+3%4 = 5%4+6%4+7%4 (2 parts = k parts)

9%4+10%4 = 1%4+2%4=1+2 (sum of first 2 numbers)

You got the idea, right :>

Took 200 to 300 ms (depend on input)

Implementation: Java

Here I’m using BigInteger because the output will exceed the long’s bound in Java. (10^17)

import java.math.BigInteger;

import java.util.Scanner;

public class EasySum {

publicstaticvoidmain(String[] args) {

Scannerinput = new Scanner(System.in);

int t = input.nextInt();

int n,m,k,remainder;

for(int i = 1; i <= t; ++i) {

n = input.nextInt();

m = input.nextInt();

k = (int) Math.floorDiv(n,m);

remainder= n%m;

m = m1;

BigIntegerbigK = BigInteger.valueOf(k);

BigIntegerbigM = BigInteger.valueOf(m);

BigIntegerbigRemainder = BigInteger.valueOf(remainder);

if(remainder == 0) {

//(long) ((m*(m+1))/2)*k;

} else {

//(long) ((m*(m+1))/2)*k + ((remainder*(remainder+1)/2));

}

}

}

}

## Problem 13: Large sum

*Difficulty: Easy

Problem:

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

How to solve:

You can do this either the hard way or the easy way. I mean if you want to save your time and don’t want to reinvent the wheel, then Java with BigInteger is the best fit for your requirement. But if you want to implement the adding mechanism yourself, then here’s how you do it.

For example, we got 2 number, 1234567 and 1234567
1234567

+1234567

let loop from the back to the top

remainder <- 0

result <- “”

7+7+remainder = 14 -> result = “4”, remainder = 1;

6+6+remainder = 13 -> result = “34”, remainder = 1;

5 +5+remainder = 11-> result = “134”, remainder = 1;

4+4+remainder = 9 -> result = “9134”, remainder = 0;

3+3+remainder = 6 -> result = “69134”, remainder = 0;

2+2+remainder = 4 -> result = “469134”, remainder = 0;

1+1+remainder = 2 -> result = “2469134”, remainder = 0;

You got the idea.

Took 8ms

Implementation: Java

import java.math.BigInteger;

public class Prob13 {

privatestaticfinalString[] 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”};

publicstaticvoidmain(String[] args) {

long cur = System.currentTimeMillis();

BigIntegerbigSum = BigInteger.ZERO;

System.out.println(numbers.length);

for(int i = 0; i < numbers.length; ++i) {

}

System.out.println(bigSum.toString(10).substring(0, 10));

System.out.println(“Took “ + (System.currentTimeMillis()cur) + ” ms”);

}

}

## Problem 12: Highly divisible triangular number

*Difficulty: Easy

Problem:

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: 1,2,4,7,14,28We 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?

How to solve:

I could only come up with two approaches. These approaches are different than each other only at the way they calculate the number of divisors of a number.
The first one use sieve Eratosthenes to create a list of prime numbers, then use that list to calculate the divisors by prime factor. The second one uses the most common way to generate all the divisors of a number. That is looping i from 1 to square root of n (n is the given number), if i is divisible by n, add 2 to the counter. Then print out the counter. But we should notice that it is not right with the perfect square number. So before we print out the counter, we check if sqrt of n to the power of 2 is equal with n , if it is, we minus 1 from counter.

The first one took 254ms, the second one took 10 time longer, about 2.5 seconds.

Implementation: Java

import java.util.Arrays;

public class Prob12 {

privatestaticboolean[] sieveEratosthenes;

privatestaticfinalintSIZE = 100000;

publicstaticvoidmain(String[] args) {

long cur = System.currentTimeMillis();

/*

int countDivisors = 0;

int triangularNumber = 1;

int ith = 1;

while(true) {

if(numberOfDivisors(triangularNumber) >= 500) {

System.out.println(triangularNumber);

System.out.println(“Took ” + (System.currentTimeMillis()-cur) + ” ms“);

//247ms

break;

}

++ith;

triangularNumber = ((ith+1)*ith)/2;

}

*/

sieveEratosthenes= new boolean[SIZE];

Arrays.fill(sieveEratosthenes, true);

for(int i = 2; i < (int)Math.sqrt(sieveEratosthenes.length); ++i) {

if(sieveEratosthenes[i])

for(int j = i*2; j < sieveEratosthenes.length; j+=i)

sieveEratosthenes[j] = false;

}

for(int i = 2; i < sieveEratosthenes.length; ++i)

int n = 1;

int i = 1;

while(true) {

if(numberOfDivisorsUsingPrimeList(primeList, n) >= 500) {

System.out.println(n);

System.out.println(“Took “ + (System.currentTimeMillis()cur) + ” ms”);

break;

}

++i;

n = ((i+1)*i)/2;

}

//http://www.mathblog.dk/triangle-number-with-more-than-500-divisors/

}

private static int numberOfDivisorsUsingPrimeList(LinkedList<Integer> primeList, int n) {

int i = 0; //first index in prime list

int divisors = 1;

int exp = 0;

while(n>1) {

int nthPrime = primeList.get(i);

if(n%nthPrime == 0) {

++exp;

n /= nthPrime;

if(n == 1) divisors*= (exp+1);

} else {

if(exp > 0) {

divisors*= (exp+1);

exp = 0;

}

++i;

}

}

return divisors;

}

privatestaticintnumberOfDivisors(int n) {

intbound = (int)Math.sqrt(n);

int divisors = 0;

for(int i = 1; i <= bound; ++i){

if(n%i==0)

divisors+=2;

}

//deal with perfect square number

if(bound*bound == n)

divisors;

return divisors;

}

}

## Problem 11: Largest product in a grid

*Difficulty: Easy

Problem:

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?

How to solve:

We calculate max product in horizontal, vertical and diagonal direction, respectively. Then check out which one is max.

Solution took 1ms

Implementation: Java

public class Prob11 {

privatestaticfinalint[][] grid = {

{ 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, 04, 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, 01, 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, 07, 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}

};

public static voidmain(String[] args) {

long cur = System.currentTimeMillis();

int max = Integer.MIN_VALUE;

int maxDiagonal = diagonalProduct();

int maxHorizontal = horizontalProduct();

int maxVertical = verticalProduct();

max= maxDiagonal;

max= Integer.max(max, maxHorizontal);

max= Integer.max(max, maxVertical);

System.out.println(max);

System.out.println(“Took “ + (System.currentTimeMillis()cur) + ” ms”);

}

private static int horizontalProduct() {

int max = Integer.MIN_VALUE;

for(int i = 0; i < grid.length; ++i) {

for(int j = 0; j < grid[i].length3; ++j) {

int product = grid[i][j]*grid[i][j+1]*grid[i][j+2]*grid[i][j+3];

if(product > max)

max= product;

}

}

return max;

}

private static int verticalProduct() {

int max = Integer.MIN_VALUE;

for(int i = 0; i < grid.length3; ++i) {

for(int j = 0; j < grid[i].length; ++j) {

int product = grid[i][j]*grid[i+1][j]*grid[i+2][j]*grid[i+3][j];

if(product > max)

max= product;

}

}

return max;

}

private static int diagonalProduct() {

int max = Integer.MIN_VALUE;

for(int i = 0; i < grid.length3; ++i) {

for(int j = 0; j < grid[i].length3; ++j) {

int product = grid[i][j]*grid[i+1][j+1]*grid[i+2][j+2]*grid[i+3][j+3];

if(product > max)

max= product;

}

}

for(int i = 0; i < grid.length3; ++i) {

for(int j = 3; j < grid[i].length; ++j ) {

int product = grid[i][j]*grid[i+1][j1]*grid[i+2][j2]*grid[i+3][j3];

if(product > max)

max= product;

}

}

return max;

}

}

## Problem 10: Summation of primes

*Difficulty: Easy

Problem:

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

Find the sum of all the primes below two million.

How to solve:

As problem 7, we have at least two ways to solve this problem.

• 1 :  Using sieve Eratosthenes to create a dictionary of prime numbers that are smaller than 2,000,000. Then, go through the dictionary and sum up all the prime numbers.
• 2:  Using a helping function to determine if a number is prime or not. The speed and run time of this method heavily depend on how you construct the helping function. Here I’m using naive primality test. You can reference pseudo code on Wiki .

About the run time, the first method takes about from 30 to 60 ms , mean while, the second method takes about 250 to 340 ms

Implementation: Java

import java.util.Arrays;

public class Prob10 {

private static boolean[] sieveEratosthenes;

private static final intSIZE = 2000000;

public static void main(String[] args) {

long cur = System.currentTimeMillis();

sieveEratosthenes= new boolean[SIZE];

Arrays.fill(sieveEratosthenes, true);

for(int i = 2; i <= (int)Math.sqrt(sieveEratosthenes.length); ++i) {

if(sieveEratosthenes[i])

for(int j = i*2; j < sieveEratosthenes.length; j+=i)

sieveEratosthenes[j] = false;

}

long sum = 0;

for(int i = 2; i < sieveEratosthenes.length; ++i)

if(sieveEratosthenes[i])

sum += i;

System.out.println(sum);

System.out.println(“Took “ + (System.currentTimeMillis() cur) + ” ms”);

/*

long sum = 0;

for(int i = 2; i < SIZE; ++i) {

if(isPrime(i))

sum +=i;

}

System.out.println(sum);

System.out.println(“Took ” + (System.currentTimeMillis()-cur) + ” ms“);

*/

}

private static boolean isPrime(int n) {

if(n <= 1) return false;

else if(n <= 3)

return true;

else if(n % 2 == 0 || n % 3 == 0)

return false;

int i = 5;

while(i*i <= n) {

if(n%i == 0 || n%(i+2) == 0)

return false;

i += 6;

}

return true;

}

}

## Problem 9 : Special Pythagorean triplet

*Difficulty: Easy

Problem:

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.

How to solve:

for a from 1 to 998

—-for b from a+1 to 999

——–for c from b+1 to 1000

————if a+b+c equals 1000 and a*a+b*b equals c*c

—————-print out abc

Took 15 ms

Implementation: Java

public class Problem9 {

public static void main(String[] args) {

for(int a = 1; a <= 998; ++a) {

for(int b = a+1; b <= 999; ++b)

for(int c = b+1; c <= 1000; ++c) {

if(a+b+c == 1000 && a*a+b*b==c*c) {

System.out.println(a + ” “ + b + ” “ + c + ” – “ + a*b*c);

}

}

}

}

} //end main

}

## Problem 8: Largest product in a series

*Difficulty: Easy

Problem

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?

How to solve:

Just loop all substrings from 0 to 1000-13 (exclusive) , convert those substrings into digits, and calculate the product of all those digits.

Implementation: Java

public class Prob8 {

private static final String input =

“73167176531330624919225119674426574742355349194934”

+ “96983520312774506326239578318016984801869478851843”

+ “85861560789112949495459501737958331952853208805511”

+ “12540698747158523863050715693290963295227443043557”

+ “66896648950445244523161731856403098711121722383113”

+ “62229893423380308135336276614282806444486645238749”

+ “30358907296290491560440772390713810515859307960866”

+ “70172427121883998797908792274921901699720888093776”

+ “65727333001053367881220235421809751254540594752243”

+ “52584907711670556013604839586446706324415722155397”

+ “53697817977846174064955149290862569321978468622482”

+ “83972241375657056057490261407972968652414535100474”

+ “82166370484403199890008895243450658541227588666881”

+ “16427171479924442928230863465674813919123162824586”

+ “17866458359124566529476545682848912883142607690042”

+ “24219022671055626321111109370544217506941658960408”

+ “07198403850962455444362981230987879927244284909188”

+ “84580156166097919133875499200524063689912560717606”

+ “05886116467109405077541002256983155200055935729725”

+ “71636269561882670428252483600823257530420752963450”;

private static final intTHIRTEEN = 13;

public static void main(String[] args) {

long cur = System.currentTimeMillis();

long max = Long.MIN_VALUE;

for(int i = 0; i < input.length()THIRTEEN; ++i) {

if(product > max) {

max= product;

}

}

System.out.println(max);

System.out.println(“Took “ + (System.currentTimeMillis()cur) + ” ms”);

}

long result = 1;

for(int i = 0; i < adjacents.length(); ++i) {

result*= digits[i];

}

return result;

}

}

Took 15 ms

## Problem 7: 10001st prime

*Difficulty: Easy

Problem:

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?

How to solve:

There are many ways to solve this problem, here are my two methods.

Method 1: Using sieve Eratosthenes to create a prime numbers dictionary , and count to the 10001st prime, print it out.

Method 2: Start from 2, using a helping function to check if a number is prime (There are many possible algorithms to check prime number as well, but here I’m using naive primality test – https://en.wikipedia.org/wiki/Primality_test) and a variable , for example called countPrimes to count how many prime numbers we’ve discovered. Every time we got a new prime number, we increase countPrime to 1, and repeat the process until countPrime reach 10001.

Both methods took <10ms

Implementation (Java)

import java.util.Arrays;

public class Prob7 {

private static boolean[] sieveEratosthenes;

public static void main(String[] args) {

long cur = System.currentTimeMillis();

sieveEratosthenes= new boolean[110000];

Arrays.fill(sieveEratosthenes, true);

for(int i = 2; i <= (int)Math.sqrt(sieveEratosthenes.length); ++i) {

if(sieveEratosthenes[i]) {

for(int j = i*2; j < sieveEratosthenes.length; j+=i)

sieveEratosthenes[j] = false;

}

}

int countPrimes = 0;

for(int i = 2; i < sieveEratosthenes.length; ++i) {

if(sieveEratosthenes[i]) ++countPrimes;

if(countPrimes == 10001) {

System.out.println(i);

System.out.println(“Took “ + (System.currentTimeMillis()cur) +”                                      ms”);

break;

}

}

//brute force

/*

int countPrimes = 1;

int n = 3;

while(true) {

if(isPrime(n))

++countPrimes;

if(countPrimes == 10001) {

System.out.println(n);

System.out.println(“Took ” + (System.currentTimeMillis()-cur) + ”                                      ms“);

break;

}

n+=2;

}

*/

}

private static boolean isPrime(int n) {

if(n <= 1) return false;

else if(n <= 3)  return true;

else if(n % 2 == 0 || n % 3 == 0return false;

int i = 5;

while(i*i <= n) {

if(n%i == 0 || n%(i+2) == 0)

return false;

i += 6;

}

return true;

}

}

## Problem 6: Sum square difference

*Difficulty: Easy

Problem:

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.

How to solve:

This problem is so damn easy.

Implementation (Java):

Solution took 1ms

public class Prob6 {

public static void main(String[] args) {

long cur = System.currentTimeMillis();

int sumSquare = 0;

int squareSum = 0;

for(int i = 1; i <= 100; ++i) {

sumSquare+= (int)Math.pow(i,2);

squareSum+= i;

}

squareSum= (int)Math.pow(squareSum, 2);

System.out.println(squareSumsumSquare);

System.out.println(“Took “ + (System.currentTimeMillis()cur) + “ms”);

}

}

## Problem 5: Smallest multiple

*Difficulty: Easy

Problem

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?

How to solve it:

As you may know, if the number is divisible by all the numbers from 1 to 20, it must first divisible by 20. With this in mind, we can come up with a pretty fast and simple solution.
Start from 20, check if the number’s divisibility , if it is, we stop and print out the result, if it not, we step forward 20 steps (as explained). This will dramatically decrease run time from hundreds of milliseconds to just a few dozens of milliseconds as we just step 1 step forward.

Implementation (Java):

public class Prob5 {

public static void main(String[] args) {

long cur = System.currentTimeMillis();

long n = 20;

while(!checkDivisibility(n)) {

n+=20;

}

System.out.println(n);

System.out.println(“Took “ + (System.currentTimeMillis() cur) +” ms”);

}

private static boolean checkDivisibility(long n) {

for(int i = 1; i <= 20; ++i)

if(n % i != 0) return false;

return true;

}

}

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