A Nice Mandelbrot Problem
by djmathman, Jan 13, 2018, 4:23 AM
oops I head off to school tomorrow for second semester junior year whee
Mandelbrot 2017-2018 wrote:
Let 
be the set containing all positive integers less than 
that are not multiples of 
. Suppose is partitioned into forty sets, each containing three of the numbers, along with one set containing the single leftover number. What is the largest possible value for the single leftover number, if each set of three numbers form an arithmetic sequence with difference 
or ?
be the set containing all positive integers less than 
that are not multiples of 
. Suppose is partitioned into forty sets, each containing three of the numbers, along with one set containing the single leftover number. What is the largest possible value for the single leftover number, if each set of three numbers form an arithmetic sequence with difference 
or ?
board with 
and 
tiles and one square uncovered. This is easily seen by taking the square in the 
row and 
column (where here the lower left corner is row 
and column 
.
and 
,
in the cell in the 
is a primitive third root of unity. Then the sum of the numbers in the cells covered by any tile is zero. But the sum of the numbers in all 
cells is ![\[(1+\omega + \cdots + \omega^{10})^2 = (1+\omega)^2 = \omega.\]](http://latex.artofproblemsolving.com/4/1/c/41cf17047d7e1b2572b924ed2121657fff95e526.png)
Hence the lone cell must have its row and column number sum to something which is 
.![[asy]
size(180);
defaultpen(linewidth(0.75));
for(int i=0;i<=11;i=i+1)
{
draw((i,0)--(i,11)^^(0,i)--(11,i));
for(int j=0;j<=10;j=j+1)
{
if ((i+j) % 3 == 1 && (i < 11))
dot((i+0.5,j+0.5));
}
}
[/asy]](http://latex.artofproblemsolving.com/4/8/1/481e5ba9e0eddac3caadda8ba5b6ed8037fd3e37.png)
![[asy]
size(180);
defaultpen(linewidth(0.75));
for(int i=0;i<=11;i=i+1)
{
draw((i,0)--(i,11)^^(0,i)--(11,i));
for(int j=0;j<=10;j=j+1)
{
if ((i+j) % 3 == 1 && (i < 11) && ((i-j) % 3 == 0))
dot((i+0.5,j+0.5));
}
}
[/asy]](http://latex.artofproblemsolving.com/4/e/d/4ed0422dbfd60f9be0423aee8d8556d9f739cb11.png)
.![[asy]
size(180);
defaultpen(linewidth(0.75));
for(int i=0;i<=11;i=i+1)
{
draw((i,0)--(i,11)^^(0,i)--(11,i));
if(i < 11)
{
draw((i+0.5,0.5)--(i+0.5,2.5));
draw((i+0.5,3.5)--(i+0.5,5.5));
}
}
draw((0.5,6.5)--(0.5,8.5)^^(1.5,6.5)--(1.5,8.5)^^(9.5,8.5)--(9.5,10.5)^^(10.5,8.5)--(10.5,10.5));
for(int i=0;i<=2;i=i+1)
{
for(int j=0;j<=1;j=j+1)
{
draw((3*i+2.5,j+6.5)--(3*i+4.5,j+6.5));
draw((3*i+0.5,j+9.5)--(3*i+2.5,j+9.5));
}
}
draw((2.5,8.5)--(4.5,8.5)^^(5.5,8.5)--(7.5,8.5));
dot((8.5,8.5));
[/asy]](http://latex.artofproblemsolving.com/9/c/e/9ce3e39b7ae92caac8ce820e0608b8a8eb35badb.png)
This post has been edited 2 times. Last edited by djmathman, Jan 13, 2018, 4:42 AM
Reason: small mistake in sol
Reason: small mistake in sol
April 2025
November 2024