Hello IMOment

We see that all elements of a nice set must be of the form ${{2}^{k}}{{5}^{l}}$ for some non-negative integers $k$ and $l$ . Moreover, ${{2}^{k}}{{5}^{l}}\le 2016$ leaves us only the following possibilities:

$\begin{matrix} l=0 & \text{and} & 0\le k\le 10 \\ l=1 & \text{and} & 0\le k\le 8 \\ l=2 & \text{and} & 0\le k\le 6 \\ l=3 & \text{and} & 0\le k\le 4 \\ l=4 & \text{and} & 0\le k\le 1 \\ \end{matrix}$

For each number of the form ${{2}^{k}}{{5}^{l}}$ we can consider the quantity $k-l$ and call it value of the number ${{2}^{k}}{{5}^{l}}$ . Observe that the value of the product of two such numbers is equal to the sum of their values. Denote by ${{T}_{j}}$ the set of all numbers listed above with value equal to $j$ .

It is easily seen that: $\left| {{T}_{-4}} \right|=1$ , $\left| {{T}_{-3}} \right|=\left| {{T}_{-2}} \right|=2$ , $\left| {{T}_{-1}} \right|=3$ , $\left| {{T}_{0}} \right|=\left| {{T}_{1}} \right|=4$ , $\left| {{T}_{2}} \right|=\left| {{T}_{3}} \right|=\left| {{T}_{4}} \right|=3$ , $\left| {{T}_{5}} \right|=\left| {{T}_{6}} \right|=\left| {{T}_{7}} \right|=2$ , $\left| {{T}_{8}} \right|=\left| {{T}_{9}} \right|=\left| {{T}_{10}} \right|=1$ and all other sets ${{T}_{j}}$ are empty.

Let us also denote ${{T}_{-}}=\bigcup\limits_{j<0}{{{T}_{j}}}$ and ${{T}_{+}}=\bigcup\limits_{j>0}{{{T}_{j}}}$ , so that $\left| {{T}_{-}} \right|=8;\ \left| {{T}_{+}} \right|=22$ .

Elements of ${{T}_{-}}$ have negative values, elements of ${{T}_{+}}$ have positive values, while all elements of ${{T}_{0}}$ have value 0.

Let $S$ be a nice set. The condition that product of its elements is a power of 10 can be restated as the condition that sum of values of its elements is 0. (We simply use the fact that number of the form ${{2}^{k}}{{5}^{l}}$ is a power of 10 if and only if $k=l$ , i.e. $k-l=0$ .) The sum of values of all numbers in ${{T}_{-}}$ is $3\cdot \left( -1 \right)+2\cdot \left( -2 \right)+2\cdot \left( -3 \right)+1\cdot \left( -4 \right)=-17$ , so the total value of $S\cap {{T}_{-}}$ is at least $-17$ , and consequently the total value of $S\cap {{T}_{+}}$ is at most 17.

By a greedy strategy we respectively take elements from ${{T}_{1}},\ {{T}_{2}},\ ...,\ {{T}_{10}}$ until their total value exceeds 17. Since $4\cdot 1+3\cdot 3+3\cdot 3=19>17$ , we cannot take all elements from ${{T}_{1}}\cup {{T}_{2}}\cup {{T}_{3}}$ , so $S\cap {{T}_{+}}$ has at most $4+3+3-1=9$ elements. On the other hand if we take all elements form ${{T}_{1}}\cup {{T}_{2}}$ , one element from ${{T}_{3}}$ and one element from ${{T}_{4}}$ , the total value of $S\cap {{T}_{+}}$ will be exactly $4\cdot 1+3\cdot 2+1\cdot 3+1\cdot 4=17$ .

Therefore the largest nice set has $\left| {{T}_{-}} \right|+\left| {{T}_{0}} \right|+4+3+1+1=21$ elements.