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# Long time since i needed help

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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 ?

May 20, 2022

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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 ?????

May 20, 2022