I believe all multiples of 28 should work. (Listing the first few operations)
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We might need some additional detail for any readers who might be less gifted. (me.)
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i claim that @AwesomeLife_Math 's claim is true.
first, looking downwards from above at the table, let the top child be A (WLOG), and label the others B,C,…,F going clockwise (again WLOG, the direction doesn’t make a difference to the argument).
consider this sequence of steps:
A performs 4m (m\in\mathbb{N}) moves
each of F, D, B perform m moves respectively
finally A performs 3m moves
notice that all except A have 3m sweets. so this initial arrangement is perfect if n-25m=3m\iff n=28m. so clearly n being any multiple of 28 leads to a perfect arrangement.
now we must see if only these work (ie. the converse, so perfect arrangement \implies 28\ |\ n).
assume to the contrary that n=28m+k leads to a perfect arrangement, where 0<k<28 (so that 28\not| \ n).
further, let the number of moves each child X plays be denoted by their corresponding lowercase letter x, meaning A plays a moves, B plays b,… F plays f.
each time a move is played by anyone, a sweet is eaten.
\implies a+b+c+d+e+f=28m+k-6x, letting x denote the number of sweets each child has at the end.
also, we can find expressions for number of sweets each child has at the end.
for example, for A:
- performs a moves, losing 4a sweets due to this.
- gains a sweet for each move that B,D,F play.
- has 28m+k sweets initially.
\iff b+d+f-4a+28m+k=x (A)
repeating this for each child gives:
- (B):a+c+e-4b=x
- (C ):b+d+f-4c=x
- (D): a+c+e-4d=x
- (E): b+d+f-4e=x
- (F): a+c+e-4f=x
(B)=(D)=(F)\iff b=d=f
(C )=(E)\iff c=e
(A)=(C )\iff 28m+k=4(a-c)\iff 4(a-c-7m)=k
so 4 \ | \ k and k=4l for 0<l<7.
from (A) we now know 3b-4a+28m+4l=x and from (A)=(C ) that c=a-(7m+l).
we also know from before that
\begin{align}
a+b+c+d+e+f&=28m+k-6x
\\\iff a+2c+3b &= 28m + 4l-6(3b-4a+28m+4l)
\\\iff a+2(a-(7m+l))+3b &= -20(7m+l) +6(4a-3b)
\\\iff a+b+6(7m+l)&=2(4a-3b)
\\\iff 6(7m+l)&=7(a-b)
\\\iff 6l=7(a-b-6m)
\end{align}
this means that, as \gcd(6,7)=1, 7 \ | \ l. however there is no integer that satisfies this for the required 0<l<7, leading to a contradiction.
therefore due to 4 \ | \ k and 7 \ | \ l, we know 28 \ | \ k. this means n=28m+k=28(m+w) for some positive integer w. both m and w are arbitrary, so we now have that arrangement is perfect \iff n is a multiple of 28.
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