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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!!!!!!
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]Step 3Now 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.
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"]Step 4So 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.
[/et_pb_tab][et_pb_tab title="Hint 5" _builder_version="4.0"]
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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