Question

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?

Answer 1

// find the nth triangular number and return it
var f = function(n) {
var g;
g = n * (n + 1) / 2;
return g;
};

// find a prime number that is >= m and return it
var prime = function(m) {
for (var j = 2; j < m; j++) {
if (m % j === 0) {
j = 1;
m++;
}
}
return m;
};
var user = prompt("How many multiples should the triangular number have? At least

...");
var n = 2;
var puppy = true;
while(puppy) {
var triangular = f(n);
if (triangular >= 1000000000) {
console.log("Pyramidal number is too big.");
puppy = false;
} //if
var powers = [];
var product = 1;

for (var i = 2; i < triangular + 1; i++) {
var p = prime(i);
if (triangular % p === 0) {
powers.push(1);
triangular /= p;
while (triangular % p === 0) {
powers[(powers.length - 1)]++;
triangular /= p;
}
} //if
else {
powers.push(0);
} //else
if (p > i) {
i = p;

}
} //for

for (var l = 0; l < powers.length; l++) {
if (powers[l] > 0) {
product *= (powers[l] + 1);
} //if
} //for
if (product >= user) {
console.log("n is " + n);
console.log("Triangular number is " + f(n));
console.log("Number of factors is " + product);
console.log(powers);


puppy = false;
} //if
else {
n++;
} // else
} //while

Answer 2

he first triangular number with over 500 factors is: 76,576,500

Answer 3

The*

Answer 4

I wish I could give myself a meddle.

Answer 5

anyone who is interested in this problem can find it online here at: http://projecteuler.net/problem=12

Answer 6

that code is for javascript and yes i did write it myself "oink oink"

Answer 7

figures out the same using perl :-

```
my @triangle_numbers = ();
foreach my $number (1 .. 99999) {
my $tn = $number*($number+1)/2;
push(@triangle_numbers, $tn);
}

foreach my $number (@triangle_numbers) {
@divisors = grep { $number % $_ == 0 } 1 .. sqrt($number);
push @divisors, map {$number == $_*$_ ? () : $number/$_} reverse @divisors;
print "$number $#divisors\n" if $#divisors > 498;
}
```

Answer 8

too cool ganeshie8!

Answer 9

ty :) your code looks great ! mine took < 6 seconds.... thinking how to optimize it further...

Answer 10

i have not learned to code in pearl yet. there are some clever math tricks to speed up the program.

Answer 11

yea most of the time is taken while grabbing all factors of a number

Answer 12

this guy can do it in under 3 seconds: http://code.jasonbhill.com/sage/project-euler-problem-12/

Answer 13

or, rather, his program can do it in under 3 seconds by remembering the previous iteration's factor

Answer 14

I wish i knew coding.

Answer 15

great place to learn coding for free is http://www.codecademy.com/

Answer 16

hez using the factors of (n+1) of previous number for n in current number... thats a brilliant idea to reduce run time !!!

Answer 17

thanks for sharing @cebroski :)

Answer 18

thank you for replying to the post @ganeshie8