Proof: Sum of the combination of the these numbers are not equal. You have a set (wallet) of 5 coins: $\{1, 5, 10, 50, 100\}$.
Now there are clearly $2^5$ subsets of this set since the decision needed to build a subset is whether to include each element or not (can be coded 0/1) and you make five such decisions $2^5$ by the multiplication rule. 
What I want to show is that no combination of elements results in the same sum. Is there a neat way to proof that?
 A: Let us assume there are two different subsets which have the same sum. Now, if there are two subsets which have the same sum, it also means that there are two disjoint subsets which have the same sum. (We get the disjoint subsets by deleting the common elements.)
Neither of these sets can contain $1$ or $5$, because if they did, the sum's last digit would be $1$, $5$ or $6$ which can't be reached by the remaining three numbers.
Now,we must divide the remaining three numbers into two subsets which have the same sum. Clearly, if $100$ is in either set, it's sum will be greater than the other set.
Now, we only have two unequal elements, $10$ and $50$ which to be put into two subsets of equal sums. This wold only be possible if $10=50$, but it isn't.
A: Since $1<5, 1+5<10, 1+5+10<50, 1+5+10+50<100$, you can prove that no two collections of coins can add up to the same value.
That's because if two collections add to the same value, then you can look at the largest coin in the two sets. If it is in both sets, then you can remove it from both sets, so assume the largest coin is only in one set.
But then the other set consists of smaller coins only, and those smaller coins can't add up to even that largest coin.
A: The number of partitions of $n$ such that each number in $\{1, 5, 10, 50, 100\} $ appears at most once  is given by the coefficient of $x^n$ in $(1+x)(1+x^5)(1+x^{10})(1+x^{50})(1+x^{100})$. When you developp you obtain :$$x^{166}+x^{165}+x^{161}+x^{160}+x^{156}+x^{155}+x^{151}+x^{150}+x^{116}+x^{115}+x^{111}+x^{110}+x^{106}+x^{105}+x^{101}+x^{100}+x^{66}+x^{65}+x^{61}+x^{60}+x^{56}+x^{55}+x^{51}+x^{50}+x^{16}+x^{15}+x^{11}+x^{10}+x^6+x^5+x+1$$
So all the reachable amounts are reachable only in one way.
