Problem 12
Solution (in Ruby)
The main aspect about this solution is finding an efficient way to calculate the number of divisors. Lets take a simple example. The number 6 has 4 divisors namely 1,2,3,6. For this number its enough we run upto the Square root of 6 which is integer floored value to be 2. 1 is divisible by 6 by 6 times, so 1 and 6 are divisors. 2 is divisble by 6 3 times, so 2 and 3 are divisors. So totally we have 4 divisors.
There is one more point to be noted here. For a square number such as 16, if we apply this formula, we will end up getting 6 divisors --> 1, 2, 4, 4, 8, 16. You may see we may calculate 4 twice. But it occurs only once for a square number. This special case is handled in the solution given below. The rest is an as-is implementation of the problem statement.
Hover here to see the solution
Cheers!!
Bragaadeesh.
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:
What is the value of the first triangle number to have over five hundred divisors?
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:1: 1We can see that 28 is the first triangle number to have over five divisors.
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
What is the value of the first triangle number to have over five hundred divisors?
Solution (in Ruby)
The main aspect about this solution is finding an efficient way to calculate the number of divisors. Lets take a simple example. The number 6 has 4 divisors namely 1,2,3,6. For this number its enough we run upto the Square root of 6 which is integer floored value to be 2. 1 is divisible by 6 by 6 times, so 1 and 6 are divisors. 2 is divisble by 6 3 times, so 2 and 3 are divisors. So totally we have 4 divisors.
There is one more point to be noted here. For a square number such as 16, if we apply this formula, we will end up getting 6 divisors --> 1, 2, 4, 4, 8, 16. You may see we may calculate 4 twice. But it occurs only once for a square number. This special case is handled in the solution given below. The rest is an as-is implementation of the problem statement.
Hover here to see the solution
Cheers!!
Bragaadeesh.
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