# Understand the problem

For how many positive integers $$n$$ does 1+2+3+4+….+n evenly divide from 6n?

(a)3. (b)5. (c)7. (d)9. (e)11

##### Source of the problem
American Mathematical Contest 2005 10A Problem 21

Number Theory

6/10

##### Suggested Book
Challenges and Thrills in Pre College Mathematics

Excursion Of Mathematics

Do you really need a hint? Try it first!

Step 1.

So after having a deep look into this problem you can see that if 1+2+3+…..+n evenly divides 6n that is $$\frac{6n}{1+2+3+….+n}$$ now to think about formula of the sum of 1+2+3+…..+n.

Step 2.

After getting the formula as 1+2+3+4+….+n=$$\frac{n(n+1)}{2}$$ substitute it in the equation $$\frac{6n}{1+2+3+….+n}$$ and simplify it. Give it a try!!!!!!

Step 3

Now by simplifying you will get $$\frac{12}{n+1}$$. Now here lies the main concept of this problem as you have to find integer n so you must see that if (n+1) is a factor of 12 then only $$\frac{12}{n+1}$$ will become an integer. Now find out the factors of 12 and try to build up some logic how to make this $$\frac{12}{n+1}$$ an integer.

Step 4

So you can easily say that the factors of 12 are 1,2,3,4,6 and 12 respectively now try to think who you can use this information here in this $$\frac{12}{n+1}$$. Like what are the values of n (from the factors of 12) in order to make it a (n+1) factor of 12.

Step 5 .

Here n can take values 0,1,2,3,5 and 11 respectively as n+1 must be a factor of 12 . But here 0 is not a positive integer so you have to exclude 0 so you are left with 5 different values of n . So your answer is 5