## Two tough questions

Tried a lot but not sure if i am correct. I dont have the right answers Q1. Find the largest positive integer n such that 1^2 + 2^2 + 3^2 + . . . +n^2 is divisible by 1+2+3 +. . . .+ n. By trail and error it seems n cant be more than 1 but is there any better way of proving it Q2. Three distinct lines are drawn in the plane. Suppose there exist exactly n circles in the plane tangent to all three lines. Find all possible values of n.

## not sure...

hii sohani..it wud be better if you provide us the options...realy i dnt noe frm where to strt da questn...at first go ya ans seems to b rite...bt ven i see ma soln..ma soln also seems to b rite.. ma way f solvin this questn is...... we noe that: 1^2+2^2+3^2+......+n^2=n(n+1)(2n+1)/6 ....(1) and 1+2+3+.........+n=n(n+1)/2 .......(2) as questn demands..divide eqn 1/2 we get: {2(n)+1}/3 so we hv to find greatest N so whole term can be divisile..dis z da pt where i get stuck...frm here onwrds options vil start wrkin... so bettr provide us da soln if ya can..... thnx

## ImpressingSorry but I

Impressing Sorry but I donâ€™t have the options but I think your solutions is more than what I was looking for Thanks

## continuing the discussion

continuing the discussion (2n+1)/3 = k = + ve integer (4n+2)/3 = n+ (n+2)/3 = 2k = k' = +ve integer thus (n+2)/3 = k'' = +integer thus n = 3k'' -2 for k'' = 1,2,.... so on seems the max number fo n is infinity to me as long as it is in the above form.

## Answer to the 2nd question

It depends if the lines are parallel or not. Please refer to the below figure when none of the lines are parallel, possible number of circles = 4 when two of the lines are parallel, possible number of circles = 2 when all lines are parallel, possible number of circles = 0

## Q1. Find the largest

Q1. Find the largest positive integer n such that 1^2 + 2^2 + 3^2 + . . . +n^2 is divisible by 1+2+3 +. . . .+ n. By trail and error it seems n cant be more than 1 but is there any better way of proving it 1^2 + 2^2 + 3^2 + . . . +n^2 = n(n+1)(2n+1)/6 and 1+2+3 +. . . .+ n = n(n+1)/2 so on divison we get the quotient as (2n+1)/3. So this must be an integer, i.e. 2n+1 must be a multiple of 3. So, 2n+1 = 3k, where k is any natural number (1,2,3,...) =>2N = 3K - 1 AND n = (3k - 1)/2 = 1, 4, 7, 10 and so on.