What is the smallest possible positive integer value of $\delta,$ such that for some integers $\alpha,$ $\beta,$ $\gamma,$ $\delta$ and $\epsilon$ , the following condition is true?

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Condition:
**
For every five-digit number
$\overline{abcde}$
that is a multiple of
$32,$
$\alpha a + \beta b + \gamma c + \delta d + \epsilon e$
is also a multiple of
$32.$

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Details and assumptions
**

$\overline{abc}$ means $100a + 10b + 1c$ , as opposed to $a \times b \times c$ . As an explicit example, for $a=2, b=3, c=4$ , $\overline{abc} = 234$ and not $2 \times 3 \times 4 = 24$ .

The answer is 2.

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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 $\boxed{2}$