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

AI Thread Summary
To determine the maximum number of elements \( n \) that can be selected from the set \( A = \{1, 2, 3, \ldots, 2015\} \) such that the sum of any three chosen numbers is a multiple of 27, it is essential to analyze the residues of the numbers modulo 27. The residues range from 0 to 26, and for the sum of any three residues to be a multiple of 27, they must be selected carefully to maintain this condition. The maximum \( n \) can be derived from the combinations of residues that satisfy the multiple of 27 requirement. The largest number that can be included in set \( B \) is also relevant in this context. The solution provides a systematic approach to finding both \( max(n) \) and the largest number in \( A \).
Albert1
Messages
1,221
Reaction score
0
$A=\begin{Bmatrix}
{1,2,3,4,5,------,2015}
\end{Bmatrix}$
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 $A$
 
Mathematics news on Phys.org
Albert said:
$A=\begin{Bmatrix}
{1,2,3,4,5,------,2015}
\end{Bmatrix}$
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 $A$

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 $n = 75$
 
kaliprasad said:
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
$B=(9,36,63,------,2007)$
we have 75 elements in $B$
$max(n)=75$
and the largest number must be taken from $A$ is $2007$
 
Last edited:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Back
Top