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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

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