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 

Difficulty Level

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


Start with hints

Watch video

Connected Program at Cheenta

Amc Training Camp

Cheenta AMC Training Camp consists of live group and one on one classes, 24/7 doubt clearing and continuous problem solving streams.

Start for free. 

Similar Problems