every integers from 1 to 121 can be written as 5 powers of 3 We have a two-pan balance and have 5 integer weights with which it is possible to weight
exactly all the weights integers from 1 to 121 Kg.The weights can be placed all on a plate but you can also put some in a dish and others with the goods to be weighted.
It's asked to find the 5 weights that give us this possibility. It also asks you to prove that the group of five weights is the only one that solves the problem.
Easily i found the 5 weights: 1 , 3 , 9 , 27 , 81
but i can' demonstrate that this group of five weights is the only one that solves the problem.
Can you help me ?  
Thanks in advance !
 A: Your five weights add to 121. If there were another set of weights, they would have to sum to at least 121. You can obtain a contradiction. If the lower four weights are the same, you need 81 to hit 121. 
Show inductively that with one weight you can get to 1, with two to 4, with three to 13, with four to 40, with five to 121.
It works as follows - there are three possibilities for each weight - in either pan or in neither. There is one possibility for the case where no weights are in either pan, which weighs zero. The remaining cases come in pairs (swap the weights between pans, they weigh the same weight). These are all distinct (the ternary expansion of a number tells you how to allocate the weights) so the maximum number for $n$ weights is $\cfrac {3^n-1}2$.
In response to comment below - now work as follows. The above shows that the set $\{1, 3, 9, 27, 81\}$  works, and that it is efficient (each weight can be weighed in just one way). Suppose we have another set $a \le b \le c \le d \le e$ which does the same. We can immediately replace $\le$ by $\lt$ throughout, because if $a=b$ our set is not efficient, and we can't weigh $121$ distinct weights.
We naturally need $a+b+c+d+e=121$ since this will be the maximum we can weigh. In order to weigh $120$ we need $a=1$. We can't have $b=2$ else we can weigh $1=a=b-a$ (this is inefficient - note, this is where it is useful to have the inductive step in place) but to weigh $118$ we then need $b=3$. This enables us to weigh up to 4. If $c<9$ then $c-a-b \le 4$, and we can weigh the same using just $a$ and $b$ - so we are not efficient (using the inductive step again). If $c>9$ we can't weigh $121-9=112$.
A: Hint: If the first 4 weights sum up to strictly less than 40, then you can not weigh either 41kg or 121 kg.

If the fifth weight is $\leq 81$, then you can't reach 121. If the fifth weight is $\geq 81$, then you can't reach 41.
A: A more general result says, given weights $w_1\le w_2 \le \dots \le w_n$, and if $S_k = \sum_{i=1}^k w_k$; $S_0 = 0$, then everything from $1$ to $S_n$ is weighable iff each of the following inequalities hold:
$$S_{k+1} \le 3S_k + 1 \text{ for } k = 0,\dots,n-1$$
Note that this is equivalent to $w_k \le 2 S_k +1$ for each $k$.
If you wish, a proof is given here. 
A: The style of numbers, based on the balanced base 3, gives these exactly.
The proof that the powers of three are the only set, can be proven by the following.  Suppose the weights are  abcde, which can be either 1 (opposite the load), 0 off the scale, or M with the load.  
The set E represents all weights reachable by the weights a-e.  This is required to be compact, but for every combination for +x, simply reversing the 1 and M gives -x.  So the compact range goes from -121 to +121, ie all posibilities.  We see that any weight by itself must be integer, ie 01000.  
Swapping a weight from 1 to M, makes an even change, so an odd change requires the removal of a weight.  When 11111 represents 121, and removal of the lightest gives 120, then we see that e=1 is the only solution.
We now represent D as a combination of a,b,c,d, and note that the 81 combinations of D become 243 of the form D+e, D, D-e, D'+e, D', D'-e, where D and D' are adjacent combinations of a,b,c,d.  Because this implies that all D must be multiples of three.
Dividing D into some minimum d and a set C.  The arguement is as above: the value of d is 120 - 117 (the largest two members), and all of C is 9.  Repeat to show that c=9, b=27 and a=81. 
Failing this, one can add the number 121 (11111 in base 3), that gives a different number, where the weight is left off the pan (1), in with the object to weigh (0), or on the side with the major weight (2).
Suppose the weight to weigh is a stone (14 lb), the number in base 3, is 112.  Adding 11111 to this, gives 12000.  This means we leave out the 81 weight (1), put the 27-weight opposite the stone, and put the 9, 3, and 1 weights with the stone.  The balanced form of 14 is 1MMM, where M is the digit for -1.  
The same number system can be used to find a faulty coin (in excess or deficit), from 120 coins, with just 5 weighings.  Number the coins from 1 to 121, excluding 61, according to the balanced rule, or from 122 to 242, excluding 182, according to the usual base-3 style numbers.  The odd coins are then reversed (ie if there are an odd number of M,1, then the 1's and M's are swapped.
The five weighings, then proceed by the digits of the number.  For the $n$ weighing, put coins with an M in the $n$ column, in the left, and those marked $1$ in the right.  If the M pan descends, write $M$, if the $1$ pan descends, write $1$.  Otherwise write $0$.  Proceed in a similar manner through all five columns, until you have a five-place number.
If there is an odd number of $M$ and $1$ (together), one needs to reverse these (to $1$ and $M$).  The number at the end of this, will tell you if the coin is in defect (the first non-zero number is an M), or excess (1),  Swapping M with 1 in all the places will reveal the number of the defect coin, the number as stands is the excess.
So a weight giving results 0,M,1,1,1  tells us that it is an even coin, so no reversal is needed.  It tells us that the coin is in defect (ie the first weight is M), and that finally it's the coin we're calling 14 (1MMM = 14).
If 1MMM is too much, one can write it as 1-centric form (12000), and subtract 121 from that (135 - 121 = 14).
