16a+8b+4c+10d+e

From the given conditions, we have $32|10000a+1000b+100c+10d+e$ . Subsequently, this means that: $32|10000a+1000b+100c+10d+e+64e$ $=10000a+1000b+100c+10d+65e$ .

Since $gcd(32,5)=1$ , we can divide the RHS by 5 to get $32|2000a+200b+20c+2d+13e$ . It now suffices to show that $2$ is the least positive integer value for $\delta$ . Just taking any number where $a,b,c,e$ are even and $d$ is odd will show that $\delta=1$ does not suffice (such as 40256)

Thus, the smallest positive integer value for $\delta$ is $\boxed{2}$

We are given $32 | 10000a + 1000b + 100c + 10d + e$ . The factors of $10$ are $2 * 5$ , and since all of the terms are divisible by $5$ except for the last one, we look to make the last one a multiple of 5. This can be done by adding $64e$ (since e is an integer, it still satisfies the divisibility condition). Since $32$ and $5$ are coprime, we can divide the right hand side by 5 while still preserving the divisibility condition, resulting in $2000a + 200b + 20c + 2d + 13e$ . So now the answer is either $1$ or $2$ . $1$ doesn't work if d is odd, e.g. abcde = 22656, as the resulting sum would be odd. Thus, $2$ is minimized as possible.

We denote $\overline{abcde}$ by $abcde$

$\textbf{1.}$ Consider $abcde = 20000 = 32*675$

Then, $32 | 2 \alpha$ and hence $16 | \alpha$ , $\alpha = 16k_1$

$\textbf{2.}$ Now let us consider $abcde = 12000 = 32*375$ .This gives us $32 | 16k+2\beta$ or $16| 2\beta$ or $8|\beta$ , hence $\beta \ge 8$ . Clearly $\beta = 8$ is possible since $32 | 10000a+1000b+100c+10d+e$ $32 | 10000a+8b+100c+10d+e$ Setting $\alpha = 1000$ , $\beta = 8$ , $\gamma$ = 100, $\delta$ = 10 & $\epsilon = 1$ ,we have our condition satisfied. So $\beta = 8$ is the smallest possible value of $\beta$ .

$\textbf{3.}$ Let us consider abcde = $24800 = 32*775$

So $32|2\alpha + 4\beta+8\gamma$ $32|32k_1 + 32k_2+8\gamma$ \ $32|8\gamma$ $4|\gamma$ So $\gamma \ge 4$ let $\gamma = 4k_3$

$\textbf{4.}$ Let us consider $abcde = 20320$ . $32|2\alpha + 3\gamma + 2\delta$ If 32 divides the whole thing then 4 also does so $4|32k_1+12k_3+2\delta$ $2|\delta$

hence $\delta \ge 2$

Now we have to show that the value $\delta = 2$ is achievable.

$\textbf{5.}$ Let us take $abcde = 20480$ . So $32|2\alpha+ 4\gamma +8\delta$

If we put $\alpha = 16k_1$ and $\gamma = 4$ and $\delta=2$ , we get $32|32k_1+16+16$ $32|32k_1+32$ This is indeed true !!!So the lowest possible value of $\delta$ is indeed achievable!!! So The minimum value of $\delta$ is $\boxed{2}$ .

10000a+1000b+100c+10d+e is divisible by 32 thus

16a+8b+4c+10d+e is divisible by 32.

now we want to minimize the coefficient of d.

To do so we multiply the expression by 13

(16a+8b+4c+10d+e)*13 is divisible by 32 which gives 16a+8b+20c+2d+13e

so we get 2