MHB Find all integers, such that ....

[lfdahl]

Find all integers, $n$, such that the set $\{1,2,3,4, ...,n\}$ can be written as the disjoint union of the subsets, $A$, $B$ and $C$ -whose sums of elements are equal.

 

[Albert1]

Find all integers, $n$, such that the set $\{1,2,3,4, ...,n\}$ can be written as the disjoint union of the subsets, $A$, $B$ and $C$ -whose sums of elements are equal.

Spoiler

$S_n=\dfrac {n(n+1)}{2}$ must be a multiple of 3, and $n>4$
if :$n=5,$$S_5=15$ and $\dfrac {15}{3}=5=$ sums of elements
we may set :A={$1,4$},B={$2,3$}, C={$5$} and all its combinations
if :$n=6,$$S_6=21$ and $\dfrac {21}{3}=7=$ sums of elements
we may set :A={$1,6$},B={$3,4$}, C={$2,5$} and all its combinations
from above we get :
case 1: $n=5,8,11,14,---=3p+2 ,p\geq 1$
case 2: $n=6,9,12,15,---=3q+3 ,q\geq 1$

 

[lfdahl]

Spoiler

$S_n=\dfrac {n(n+1)}{2}$ must be a multiple of 3, and $n>4$
if :$n=5,$$S_5=15$ and $\dfrac {15}{3}=5=$ sums of elements
we may set :{$A=1,4$},{$B=2,3$}, {$C=5$} and all its combinations
if :$n=6,$$S_6=21$ and $\dfrac {21}{3}=7=$ sums of elements
we may set :{$A=1,6$},{$B=3,4$}, {$C=2,5$} and all its combinations
from above we get :
case 1: $n=5,8,11,14,---=3p+2 ,p\geq 1$
case 2: $n=6,9,12,15,---=3q+3 ,q\geq 1$

Good job, Albert!Thankyou for your solution!

 

[kaliprasad]

Solution provided by Albert is good but can we make the sets A, B, C for the above n. he has provided for some cases but not all.
I here mention how to build A,B,C. this is one method and takes care of solution and not all solutions
Now for the common part

Spoiler

Any six consecutive number say 6p to 6p+5 we can choose $ A = \{6p, 6p+5\} , B = \{6p + 1, 6p+4\} ,C = \{6p+2, 6p+3\}$
1st 9 numbers we can choose $A = \{4,9,2\},B = \{3,5,7\}, C =\{(1,6,8\}$

using above

Spoiler

for the case of n = n is of the form 3p + 2 where p >= 1

p =1 or n= 5 gives $\{4,1\},\{2,3\}, \{5\}$ ( Albert has mentioned)

p =2 or n= 8 gives $\{4,8\},\{1,5,6\}, \{(2,3,7\}$

for p >2 we if p odd we have n = 5 + 6k
we get the groups above( n = 5) and from groups of 6 as mentioned above

if p even we have n = 8 + 6k
we get the groups above( n = 8) and from groups of 6 as mentioned above

for n of the form 3p:

p is even: we can choose the elements from groups of 6 then combine for example to choose from (1..12) choose ((1,6),(2,5),(3,4)) from (1.. 6) and (7,12), (8,11),(9, 10) from (7..12) and combine to get (1,6,7,12),(2,5,8,11),(3,4,9,10)

p is odd: choose from 1st 9 elements and elements from groups of 6 and combine to get the result

 

[lfdahl]

Hi, kaliprasad and Albert. Thankyou for your participation and clever solutions!

Your discussion on the possible values of $n$ is interesting, and the suggested solution below uses a surprisingly short and clear argument:

Spoiler

To find the possible $n$-values, observe that:$\sum_{x \in A}+\sum_{x \in B}+\sum_{x \in C} = \frac{1}{2}n(n+1)$, which is divisible by $3$.It follows, that $n$ must be congruent to $0,2,3$ or $5$ modulo $6$. Therefore $n \le 4$ can be excluded.So the list of $n$-values begins with: $n \in \left\{5,6,8,9,11, ...\right\}$For $n = 5,6,8,9$ we have the following partitions:\[ n = 5:\: \: \: \: A = \left \{ 1,4 \right \}\: \: \: B = \left \{ 2,3 \right \}\: \: \: C = \left \{ 5 \right \} \\\\ n = 6:\: \: \: \: A = \left \{ 1,6 \right \}\: \: \: B = \left \{2,5 \right \}\: \: \: C = \left \{3,4 \right \} \\\\ n = 8:\: \: \: \: A = \left \{1,2,3,6 \right \}\: \: \: B = \left \{ 5,7 \right \}\: \: \: C = \left \{ 4,8 \right \} \\\\ n = 9:\: \: \: \: A = \left \{ 1,2,3,4,5 \right \}\: \: \: B = \left \{7,8 \right \}\: \: \: C = \left \{ 6,9 \right \} \\\\ \]Now, to proceed with a $n+6$-value (e.g. $11$), we can use the nice procedure, which kaliprasad suggested:Use the partition depicted for $n=5$: Now join $n+1$ and $n+6$ to $A$, $n+2$ and $n+5$ to $B$ - and $n+3$ and $n+4$ to $C$.

 

[Albert1]

Hi, kaliprasad and Albert. Thankyou for your participation and clever solutions!

Your discussion on the possible values of $n$ is interesting, and the suggested solution below uses a surprisingly short and clear argument:

Spoiler

To find the possible $n$-values, observe that:$\sum_{x \in A}+\sum_{x \in B}+\sum_{x \in C} = \frac{1}{2}n(n+1)$, which is divisible by $3$.It follows, that $n$ must be congruent to $0,2,3$ or $5$ modulo $6$. Therefore $n \le 4$ can be excluded.So the list of $n$-values begins with: $n \in \left\{5,6,8,9,11, ...\right\}$For $n = 5,6,8,9$ we have the following partitions:\[ n = 5:\: \: \: \: A = \left \{ 1,4 \right \}\: \: \: B = \left \{ 2,3 \right \}\: \: \: C = \left \{ 5 \right \} \\\\ n = 6:\: \: \: \: A = \left \{ 1,6 \right \}\: \: \: B = \left \{2,5 \right \}\: \: \: C = \left \{3,4 \right \} \\\\ n = 8:\: \: \: \: A = \left \{1,2,3,6 \right \}\: \: \: B = \left \{ 5,7 \right \}\: \: \: C = \left \{ 4,8 \right \} \\\\ n = 9:\: \: \: \: A = \left \{ 1,2,3,4,5 \right \}\: \: \: B = \left \{7,8 \right \}\: \: \: C = \left \{ 6,9 \right \} \\\\ \]Now, to proceed with a $n+6$-value (e.g. $11$), we can use the nice procedure, which kaliprasad suggested:Use the partition depicted for $n=5$: Now join $n+1$ and $n+6$ to $A$, $n+2$ and $n+5$ to $B$ - and $n+3$ and $n+4$ to $C$.

Spoiler

this is a question of combination
in fact $A,B,\,\,and\,\, C$ are interchangeable
for $n=5,8,11,14, ----=3p+2, p\in N, P\geq 1 $ in one group
and $n=6,9,12,15----=3q+3 ,q\in N, q\geq 1 $ as another category
for each type of n,using method of combinatin to find all possible combinatons of $A,B,$and $C$
as mention above
in this question $A,B,$ and $C$ are not main characters, for we are asked to find all possible values of n
take $n = 8:\: \: \: \: A = \left \{4,8 \right \}\: \: \: B = \left \{ 1,5,6 \right \}\: \: \: C = \left \{ 2,3,7\right \}$
also we may set $A = \left \{5,7 \right \}\: \: \: B = \left \{ 4,8 \right \}\: \: \: C = \left \{ 1,2,3,6\right \}$
so the sets of $A,B,C$ are not unique