+83 For all positive integers n, the nth triangular number t(n) is defined as 1+2+3+4+5+....+n. What is the greatest possible value of the greatest common divisor of 4(t(n)) and n-1 ?
+118609 Hi Atlas,
I don't know the answer, I am just playing with the question.
We have \(t_n=\frac{n(n+1)}{2}\)
so \(4t_n=2n(n+1)\)
the questions says what is the greatest possible value of the GCD of 2n(n+1) and n-1
If n=3 then the greatest divisor is 2
If n=5 then the greatest divisor is 4
I don't know what the biggest one is though. Maybe 4 ?????
