Let total work be measured in system-days.

If all $m$ systems worked throughout, the assignment would finish in $17$ days. Hence total work is $W = 17m.$

Now account for the crashes.

• On day 1: all $m$ systems work.
• On day 2: $4$ systems have crashed, so systems working $= m-4$.
• On day 3: $8$ systems have crashed in total, so systems working $= m-8$.
• and so on.

Since it took $8$ more days, total time becomes $17+8 = 25 \text{ days}.$

Therefore, the number of working systems each day forms an AP: $m,\; m-4,\; m-8,\; \dots$ for $25$ days.

Find the total work done in these 25 days.

This is the sum of first $25$ terms of the AP with $a = m, \quad d = -4.$

So, $S_{25} = \frac{25}{2}\left[2m + (25-1)(-4)\right].$

$S_{25} = \frac{25}{2}(2m - 96).$

This total work must equal the original work $17m$: $\frac{25}{2}(2m - 96) = 17m.$

Solve for $m$.

Multiply both sides by $2$: $25(2m - 96) = 34m$ $50m - 2400 = 34m$ $16m = 2400$ $m = 150.$

Check with options.

$m=150$ corresponds to Option C.

Comparison with stored answer.

Stored correct answer is C, which matches our result.