Show that all the cards contain the same number. Natural numbers from $1$ to $99$ (not necessarily distinct) are written on $99$ cards. It is given that the sum of the numbers on any subset of cards (including the set of all cards) is not divisible by $100$. Show that all the cards contain the same number.
I was suggested by a friend to approach this problem by contradiction.
Assume to the contrary that there are two cards which are distinct. 
However, I am not really seeing how that would help.
Does anyone see how this proof would be approached? Any starting point would help me a lot.
 A: Let the number on the cards be $\{a_1,a_2, \ldots , a_{99}\}$. Suppose $a_1 \neq a_2$. Then consider the following $99$ subsets
$$\{a_1\}, \{a_1,a_2\}, \{a_1,a_2,a_3\}, \ldots \{a_1,a_2, \ldots a_{99}\}.$$
Then the corresponding sums will be
 $$s_1=a_1,\, s_2=a_1+a_2, \, s_3=a_1+a_2+a_3, \ldots , \, s_{99}=a_1+a_2+ \dotsb + a_{99}.$$
Since none of them is divisible by $100$, therefore modulo $100$ they should all give distinct remainders (and also cover all possible non-zero remainders modulo $100$). If not, then there exists $i,j$ with $i < j$ such that
$s_i \equiv s_j \pmod{100}.$ But then the subset $\{a_{i+1}, a_{i+2}, \ldots ,a_{j}\}$ will have sum $s_j-s_i$ and it will be divisible by $100$ contradicting the condition given.
Now consider $a_2$. Since $a_1,a_2 \in \{1,2,\ldots ,99\}$ and are distinct therefore $a_2 \not\equiv a_1 \not\equiv s_1\pmod{100}$. Also $a_2 \not\equiv s_2 \pmod{100}$ either, otherwise $a_1 \equiv 0 \pmod{100}$.
Therefore $a_2 \equiv s_k \pmod{100}$ for some $k \geq 3$.
This is a contradiction because then the subset $\{a_1,a_3, \ldots ,a_{k}\}$ will have the sum divisible by $100$.
So we can only have $a_1=a_2$. 
