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

Topic
Number Theory

Difficulty Level
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

 

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