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?

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

n → n/2 (n is even)
n → 3 n + 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.

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

Find the sum of all the primes below two million.

/**
 *
 * @author olyjosh
 */
public class SumOfPrime {
    public static void main(String[] args) {
        int max = 2000000;

        //I implemented Sieve by taking true to be false n vice-versa, Just to avoid initial looping/flagging bla-bla
        boolean[] isPrime = new boolean[max];
        for (int i = 2; i*i < max; i++) {
            if (!isPrime[i]) {
                for (int j = i; i*j < max; j++) {
                    isPrime[i*j] = true;
                }
            }
        }

        long sum = 0;
        for (int i = 2; i < max; i++) {
            if (!isPrime[i]) sum+=i;
        }
        System.out.println("Sum of prime <" + max + " is " + sum);
    }
}

Sum of prime <2000000 is 142913828922

kudaisi:
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?

Editing ...

olyjosh:


Cool implementation bro. But the best approach to this isn't greedy approach. The sum arithmetic progression is what you can use in both case. These are formula you can easily derive using mathematical inductions.

Sn = {n(1+n)}/2
Sum of squres in AP is = {n(n+1)(2*n+1)}/6
where one 1 can alway be the first term if your series does not start from 1 and n is the last term

I'm actually still relatively new to all this stuff. My knowledge isn't as vast as yours and that was the best implementation I could come up with. I'll also be lying if I should claim to understand this formula you talk about.

kudaisi:
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?

Ans: 76576500 is the first triangle number with over 500 divisors, 576 divisors to be exact
Suboptimal bruteforce. . .. Could significantly improve it by sieving and/or making informed speculations

#include <iostream>
#include <cmath>
using namespace std;
 
int main() {
	std::size_t val{1}, idx{1}, divisors{0};
	for(; divisors < 500; idx++, val += idx){
		divisors = 0;
		for(std::size_t i{1}; i < std::size_t(std::sqrt(val)); i++)
			if(val % i == 0)
				divisors += 2;
	}
	std::cout << "Ans: " << val - idx << "\nhas " << divisors << " divisors\n";
	return 0;
}

WhiZTiM:


Ans: 76576500 is the first triangle number with over 500 divisors, 576 divisors to be exact
Suboptimal bruteforce. . .. Could significantly improve it by sieving and/or making informed speculations

#include <iostream>
#include <cmath>
using namespace std;
 
int main() {
	std::size_t val{1}, idx{1}, divisors{0};
	for(; divisors < 500; idx++, val += idx){
		divisors = 0;
		for(std::size_t i{1}; i < std::size_t(std::sqrt(val)); i++)
			if(val % i == 0)
				divisors += 2;
	}
	std::cout << "Ans: " << val - idx << "\nhas " << divisors << " divisors\n";
	return 0;
}

which language is this?

Singapore1:
which language is this?

c++

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

n → n/2 (n is even)
n → 3 n + 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.

Ans: Starting number: 837799 with length: 524
Runtime 0.01sec; (with memorization) 30x speedup... ~10,000 more memory used


But interestingly,
I just began tinkering on this problem, and I thought, "Let me do something stupid once again, bruteforce"... so I implemented it in several Languages... C++ Compilers excel pretty well in optimization.
Except for C++, All the other languages ran for unacceptable timings.

Never do this out there!

#include <iostream>
auto main() -> int
{
    std::size_t z{0}, num{1}, lnum{1};
    for(; num < 1'000'000; num++){
        std::size_t m = 0;
        for(auto n(num); n != 1; n = n%2 == 0 ? n/2 : n*3+1, ++m);
        if(m > z) { lnum = num; z = m; }
    }
    std::cout << "Starting number: " << lnum << " with length: " << z << '\n';
}

Ans: Starting number: 837799 with length: 524
Runtime 0.3sec;

Will be retiring now..... ....My engagements have drastically increased

Thanks Kudasi for these simple questions... ...I learned some things...
...I am a two time participant in one of the Regionals of the Most prestigious Programming contest in the World, ACM-ICPC. Participated in the ACM-ICPC South African Regionals back in 2012 and 2013.
Needless to say.

- In 2012, my team (ATBU) came 2nd in Nigeria but 40th in Africa. Covenant University's C++ team came 1st in Nigeria but 38th in Africa.
http://icpc.baylor.edu/regionals/finder/south-africa-2012/standings

- In 2013, my team (ATBU) came 1st in Nigeria but 5th in Africa. Covenant University's C++ team came 2nd in Nigeria but 18th in Africa. We would have landed top 2 if I and my partner didn't screw up some things (We correctly attempted 6 questions, but successfully submitted 3)... Since then, I haven't done much of Competitive programming.
http://icpc.baylor.edu/regionals/finder/south-africa-2013/standings

We were honored and rewarded ...
I will find time to write on that someday here on Nairaland...

olyjosh. Nice work. Keep it up bro!
lordZOUGA. ...err... I see you. ...

olyjosh:

c++

am currently learning c
and this show c++ will be piece of cake sha

Just coming back from exams. Back here to top up my IQ.

WhiZTiM:
Will be retiring now..... ....My engagements have drastically increased
Thanks Kudasi for these simple questions... ...I learned some things...
...I am a two time participant in one of the Regionals of the Most prestigious Programming contest in the World, ACM-ICPC. Participated in the ACM-ICPC South African Regionals back in 2012 and 2013.
Needless to say.
- In 2012, my team (ATBU) came 2nd in Nigeria but 40th in Africa. Covenant University's C++ team came 1st in Nigeria but 38th in Africa.
http://icpc.baylor.edu/regionals/finder/south-africa-2012/standings
- In 2013, my team (ATBU) came 1st in Nigeria but 5th in Africa. Covenant University's C++ team came 2nd in Nigeria but 18th in Africa. We would have landed top 2 if I and my partner didn't screw up some things (We correctly attempted 6 questions, but successfully submitted 3)... Since then, I haven't done much of Competitive programming.
http://icpc.baylor.edu/regionals/finder/south-africa-2013/standings

You do too much bro You are not simple men you know I never do much contest sha oo. Just once and the we are on the winning team at 2013 NACOSS Programming context(Team FUTMinna). 2014 was intern period, so let go 2015 to do something. @Whiztim Seriously, I'm inspired.

Need to reach you on watsapp - 08061662025

See image for question.

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?


NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.

kudaisi:
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?


NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.

This is my solution. Shout out to all my
bosses above. Hope to improve with time.

 
package problems;

import java.util.HashMap;
import java.util.Random;

/**
 */
class Problems {

    static HashMap<Integer, String> numberMap = new HashMap();
    static StringBuffer builder = new StringBuffer();
    static {
        numberMap.put(1, "one";
        numberMap.put(2, "two";
        numberMap.put(3, "three";
        numberMap.put(4, "four";
        numberMap.put(5, "five";
        numberMap.put(6, "six";
        numberMap.put(7, "seven";
        numberMap.put(8, "eight";
        numberMap.put(9, "nine";
        numberMap.put(10, "10";
        numberMap.put(11, "eleven";
        numberMap.put(12, "twelve";
        numberMap.put(13, "thirteen";
        numberMap.put(14, "fourteen";
        numberMap.put(15, "fifteen";
        numberMap.put(16, "sixteen";
        numberMap.put(17, "seventeen";
        numberMap.put(18, "eighteen";
        numberMap.put(19, "nineteen";
        numberMap.put(20, "twenty";
        numberMap.put(30, "thirty";
        numberMap.put(40, "forty";
        numberMap.put(50, "fifty";
        numberMap.put(60, "sixty";
        numberMap.put(70, "seventy";
        numberMap.put(80, "eighty";
        numberMap.put(90, "ninety";
        numberMap.put(100, "hundred";
        numberMap.put(1000, "thousand";

    }

    private static String noToWordsNoSpaces(int number) {
        builder.delete(0, builder.capacity()-1);
        int rem = number;
        int quotient = 0;
        //small numbers
        if (number <= 20) {
            return numberMap.get(number);
        } else if (number < 100) {
            quotient = rem / 10;
            rem = rem % 10;
            return String.format("%s%s", numberMap.get(quotient * 10), ((rem > 0) ? numberMap.get(rem) : "");
        }
        //larger numbers
        quotient = rem / 1000;
        rem = rem % 1000;
        if (quotient > 0) {
            builder.append(String.format("%s%s", numberMap.get(quotient), numberMap.get(1000)));
        }
        quotient = rem / 100;
        rem = rem % 100;
        if (quotient > 0) {
            builder.append(String.format("%s%s", numberMap.get(quotient), numberMap.get(100)));
        }
        if (rem == 0) {
            return builder.toString();
        } else {
            if (rem <= 20) {
                builder.append(String.format("and%s", numberMap.get(rem)));
            } else {
                quotient = rem / 10;
                rem = rem % 10;
                builder.append(String.format("and%s%s", numberMap.get(quotient * 10), ((rem > 0) ? numberMap.get(rem) : ""));
            }
        }
        return builder.toString();
    }

    public static void main(String[] args) {
        int answer = 0;
        for (int i = 1; i <= 5; i++) {
            answer += noToWordsNoSpaces(i).length();
        }
        System.out.println(answer);
        answer = 0;
        for (int i = 1; i <= 1000; i++) {
            answer += noToWordsNoSpaces(i).length();
        }
        System.out.println(answer);
    }
}

Answer is 21114.

kudaisi:
See image for question.

import java.math.BigInteger;

/**
 *
 * @author olyjosh
 */
public class SumOfDigits {
    
    public static void main(String[] args) {
        char[] n = new BigInteger("2".pow(1000).toString().toCharArray();
        int ans=0;
        for (int i = 0; i <n.length ; i++) {
            ans+=Integer.parseInt(String.valueOf(n[i]));
        }
        System.out.println("Sum of digits of number 2^1000 is "+ans);
    }
}

Sum of digits of number 2^1000 is 1366

See Image for question.

You are given the following information, but you may prefer to do some research for yourself.

1 Jan 1900 was a Monday.
Thirty days has September,
April, June and November.
All the rest have thirty-one,
Saving February alone,
Which has twenty-eight, rain or shine.
And on leap years, twenty-nine.
A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.
How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

kudaisi:
You are given the following information, but you may prefer to do some research for yourself.

1 Jan 1900 was a Monday.
Thirty days has September,
April, June and November.
All the rest have thirty-one,
Saving February alone,
Which has twenty-eight, rain or shine.
And on leap years, twenty-nine.
A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.
How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

/** @author olyjosh */
public class _01Sundays {
    
    public static int year=1901;
    
    public enum Month {
        JAN(31),FEB(28),MAR(31),APR(30),MAY(31),JUN(30),JUL(31),AUG(31),SEP(30),OCT(31),NOV(30),DEC(31);

        int days;
        private Month(int days) {
            this.days=days;
        }

    }
        
    public static void main(String []args){
        byte day =3;  //first day of jan 1901 will be deinately be tuesday
        int count=0;
        while(year<=2000){
            //iterating over month
            for(Month e :Month.values()){
                int days=e.equals(Month.FEB) && ((year%4==0 && year%100!=0) || (year%400==0)) ? 29 : e.days;
                for (int i = 1; i <= days; i++) {
                    if(day==day=1;
                    if(i==1 && day==1){
                        count++;
                    }
                    day++;
                }
            }
            year++;
        }
        System.out.println(count+" Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)";
    }
    
}

171 Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)

olyjosh:

/** @author olyjosh */
public class _01Sundays {
    
    public static int year=1901;
    
    public enum Month {
        JAN(31),FEB(28),MAR(31),APR(30),MAY(31),JUN(30),JUL(31),AUG(31),SEP(30),OCT(31),NOV(30),DEC(31);

        int days;
        private Month(int days) {
            this.days=days;
        }

    }
        
    public static void main(String []args){
        byte day =3;  //first day of jan 1901 will be deinately be tuesday
        int count=0;
        while(year<=2000){
            //iterating over month
            for(Month e :Month.values()){
                int days=e.equals(Month.FEB) && ((year%4==0 && year%100!=0) || (year%400==0)) ? 29 : e.days;
                for (int i = 1; i <= days; i++) {
                    if(day==day=1;
                    if(i==1 && day==1){
                        count++;
                    }
                    day++;
                }
            }
            year++;
        }
        System.out.println(count+" Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)";
    }
    
}

171 Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)

I think Calendar or Gregorian calendar should be able work in java anyway. I may post another solution using Calendar

olyjosh:
Just coming back from exams. Back here to top up my IQ.

You do too much bro You are not simple men you know I never do much contest sha oo. Just once and the we are on the winning team at 2013 NACOSS Programming context(Team FUTMinna). 2014 was intern period, so let go 2015 to do something. @Whiztim Seriously, I'm inspired.

Need to reach you on watsapp - 08061662025

...lol. Yes sir....
Cool!... I respect FUTMinna people. at least some of their Computer Sciences geeks.
I just saved your number, I will message you on WhatsApp.

------------------------

nnasino:

This is my solution. Shout out to all my
bosses above. Hope to improve with time.
Answer is 21114.

Mehn... Nice solution!

kudaisi:
See Image for question.

The 2x2 grid question.....

template<std::size_t M, std::size_t N>
unsigned long long count_paths(){
    unsigned long long grid[M+1][N+1] = {0};
    
    for(unsigned k = 0; k <= M; k++) grid[k][N] = 1;
    for(unsigned k = 0; k <= N; k++) grid[M][k] = 1;
    
    for(int x = M - 1; x >= 0; x--){
	for(int y = N - 1; y >= 0; y--){
	    grid[x][y] = grid[x][y+1] + grid[x+1][y];
	}
    }
    
    return grid[0][0];
}

auto main() -> void {
    std::cout << count_paths<20,20>() << '\n';
}

runs in 0.0000041seconds.... couldn't help it :-)

WhiZTiM:
Will be retiring now..... ....My engagements have drastically increased

Thanks Kudasi for these simple questions... ...I learned some things...
...I am a two time participant in one of the Regionals of the Most prestigious Programming contest in the World, ACM-ICPC. Participated in the ACM-ICPC South African Regionals back in 2012 and 2013.
Needless to say.

- In 2012, my team (ATBU) came 2nd in Nigeria but 40th in Africa. Covenant University's C++ team came 1st in Nigeria but 38th in Africa.
http://icpc.baylor.edu/regionals/finder/south-africa-2012/standings

- In 2013, my team (ATBU) came 1st in Nigeria but 5th in Africa. Covenant University's C++ team came 2nd in Nigeria but 18th in Africa. We would have landed top 2 if I and my partner didn't screw up some things (We correctly attempted 6 questions, but successfully submitted 3)... Since then, I haven't done much of Competitive programming.
http://icpc.baylor.edu/regionals/finder/south-africa-2013/standings

We were honored and rewarded ...
I will find time to write on that someday here on Nairaland...

olyjosh. Nice work. Keep it up bro!
lordZOUGA. ...err... I see you. ...

Wow boss great achievements up there. I've always been a fan of competitive programming. I tried the facebook hacker cup early this year but my solutions to the three problems were all wrong. They worked for the test cases and I was feeling fly but at the end of the round I failed all problems. Same for codejam some months later. I want to all sir for some resources I can study to make me better at optimization / algorithms /problem solving. I read CLRS by cormen et al but I would like you to point me to any good resources that can be useful to a beginner (books, websites, courses) .
Thanks for your reply in advance.

Nice job u guyz are doing. But instead of writing raw code why not give algorithm for solving the problems instead. Ive solved about 20 of these problems and I think looking through an algorithm would be much more beneficial than the actual code.

Please I need help with my algorithm. Will someone direct me? I will be really grateful.

Bonjelomo:
Please I need help with my algorithm. Will someone direct me? I will be really grateful.

We are willing to help

Jelo4kul:


We are willing to help

@jelo4cul sir, I am a newbie Java programmer. I'd love to learn algorithms too. Please put me through.

Bonjelomo:


@jelo4cul sir, I am a newbie Java programmer. I'd love to learn algorithms too. Please put me through.

As for algorithm, you just need to be good at thinking. The more complex d problem is, the more you think. As for d java programming there are many pdfs nd videos you can download online except if you want a personal tutor.

Bonjelomo:


@jelo4cul sir, I am a newbie Java programmer. I'd love to learn algorithms too. Please put me through.

As for algorithm, you just need to be good at thinking. The more complex d problem is, the more you think. As for d java programming there are many pdfs nd videos you can download online except if you want a personal tutor. But its good you learn on your own, refer to good developers to learn more.

Jelo4kul:


As for algorithm, you just need to be good at thinking. The more complex d problem is, the more you think. As for d java programming there are many pdfs nd videos you can download online except if you want a personal tutor. But its good you learn on your own, refer to good developers to learn more.

Thanks sir.

I will always come here to learn

See attachment for question:

Who are those top 10 guyz in the nigerian rank on project euler? Especially that number1 themark626