6. Let $a, b, c, d$ be odd numbers, $0<a<b<c<d$, and $a d=b c, a+d=2^{k}, b+c=2^{m}, k, m$ be integers, find the value of $a$.


6. Solution: The value of $a$ is 1.

According to the problem, $(a+d)^{2}=(a-d)^{2}+4 a d>(b-c)^{2}+4 b c=(b+c)^{2}$,
thus $a+d>b+c$, which means $k>m$.
Also, since $a\left(2^{k}-a\right)=b\left(2^{m}-b\right)$,
it follows that $(b+a)(b-a)=2^{m}\left(b-2^{k-m} a\right)$,
so $2^{m} \mid(b+a)(b-a)$.
Since $a, b$ are both odd numbers, $b+a$ and $b-a$ are both even, but their difference $2a$ is not a multiple of 4, so one of $b+a$ and $b-a$ is not divisible by 4.
Therefore, $2^{m-1} \mid(b+a)$ or $2^{m-1} \mid(b-a)$.
Noting that $0<b-a<b<\frac{1}{2}(b+c)=2^{m-1}$,
and $b+a<b+c=2^{m}$,
it follows that $2^{m-1} \mid(b+a)$ and $b+a=2^{m-1}$,
thus $b-a=2\left(b-2^{k-m} a\right)$,
leading to $a=2^{2 m-k-2}$. Since $a$ is odd,
it must be that $a=1$.