MHB Max n for Sum of 3 Numbers Multiple of 27 in A

[Albert1]

$A=\begin{Bmatrix}
{1,2,3,4,5,------,2015}
\end{Bmatrix}$
if we pick $n$ numbers from $A$, we call it the set $B$ ,and the sum of any three numbers from $B$
are multiple of 27 ,find $max(n)$ , and the largest number we can choose from $A$

 

[kaliprasad]

$A=\begin{Bmatrix}
{1,2,3,4,5,------,2015}
\end{Bmatrix}$
if we pick $n$ numbers from $A$, we call it the set $B$ ,and the sum of any three numbers from $B$
are multiple of 27 ,find $max(n)$ , and the largest number we can choose from $A$

Spoiler

sum of any 3 divisible by 27. So the numbers have to be 0 mod 27 or 9 mod 27 or 18 mod 27. This is so because all 3 have to same mod 27.
$2015 = 27 * 74 + 17$ if we take $27k + 9$ then k goes from 0 to 74 that is n = 75
27 k means 74 numbers and 27k + 18 means 74
so $n = 75$

 

[Albert1]

Spoiler

sum of any 3 divisible by 27. So the numbers have to be 0 mod 27 or 9 mod 27 or 18 mod 27. This is so because all 3 have to same mod 27.
$2015 = 27 * 74 + 17$ if we take $27k + 9$ then k goes from 0 to 74 that is n = 75
27 k means 74 numbers and 27k + 18 means 74
so $n = 75$

very good ! your answer is correct

Spoiler

$B=(9,36,63,------,2007)$
we have 75 elements in $B$
$max(n)=75$
and the largest number must be taken from $A$ is $2007$