Elementary number theory in sets

Make a following sets:

$$ A= \{1,2,4,8,16,32\}$$ $$ B= \{3,6,12,24\}$$ $$ C = \{5,10,20,40\}$$ $$ D = \{7,14,28\}$$ $$ E = \{9,18,36\}$$ $$F = \{11,22\}$$ $$G = \{13,26\}$$ $$H= \{15,30\}$$ $$I = \{17,34\}$$ $$J = \{19,38\}$$ $$K = \{21,23,25,27,29,31,33,35,37,39\}$$

Suppose the statement is not true. Then we take from each set A,B,...,J at most one element and if we take all elements from K we have a total of 20 elements. A contradiction.

Every positive integer $n$ can be written uniquely as an odd number times an integer power of two. In other words, the factorization $n=k\cdot2^i$ is unique (where $k$ is a positive odd integer and $i$ is a nonnegative integer).

When the numbers in $\{1,2,\dots,40\}$ are written in this form, the value of $k$ will be one of the $20$ positive odd numbers less than $40$. If $21$ numbers are chosen from $\{1,2,\dots,40\}$, two of them will have the form $k\cdot2^i$ for the same value of $k$, and one of those two (the one with the smaller power of $2$) will divide the other.