 # 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