A can complete a piece of work alone in 120 hours, B can complete it alone in 180 hours, while C can complete it alone in 216 hours. In every cycle of three days, only A works on Day 1, only B works on Day 2, and only C works on Day 3. In the $n^{th}$ cycle, each of A, B, and C work for $n$ hours. An exception is made on the last day, if the work is completed mid cycle. What is the total number of days needed for the work to be completed? (Note: if on the final day of work, the person working is supposed to work for p hours, but the work gets over after just q hours, that should be considered as only $\frac{q}{p}$ day.)

Total Work ($W$) = $LCM(120, 180, 216) = 1080$ units.

The work done per hour (efficiency) for each person:

A's rate: $\frac{1080}{120} = 9$ units/hour

B's rate: $\frac{1080}{180} = 6$ units/hour

C's rate: $\frac{1080}{216} = 5$ units/hour

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The work follows a cycle of 3 days. In the $n^{th}$ cycle, each person works for $n$ hours on their specific day.

Day 1 (A): Works $n$ hours $\rightarrow$ $9 \times n$ units

Day 2 (B): Works $n$ hours $\rightarrow$ $6 \times n$ units

Day 3 (C): Works $n$ hours $\rightarrow$ $5 \times n$ units

Total work done in the $n^{th}$ cycle ($W_n$):

$W_n = 9n + 6n + 5n$

$W_n = 20n \text{ units}$

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Finding how many full cycles ($k$) occur before approaching the total work of 1080 units. The cumulative work after $k$ cycles is the sum of $20n$ for $n = 1$ to $k$.

$\text{Total Work} = \sum_{n=1}^{k} 20n$

$= 20 \times \frac{k(k+1)}{2}$

$= 10k(k+1)$

Finding the largest integer $k$ such that $10k(k+1) \le 1080$, or $k(k+1) \le 108$.

If $k=9$: $9(10) = 90$ (which is $\le 108$)

If $k=10$: $10(11) = 110$ (which is $> 108$)

So, 9 full cycles are completed.

Work done in 9 cycles: $10 \times 9 \times 10 = 900$ units

Days elapsed: $9 \text{ cycles} \times 3 \text{ days/cycle} = 27$ days

Remaining Work: $1080 - 900 = 180$ units

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Now in Cycle 10 ($n=10$). The scheduled working hours for this cycle are 10 hours per person.

Day 28 (Cycle 10, Day 1): A works

Capacity: $10 \text{ hours} \times 9 \text{ units/hr} = 90$ units

Required: 180 units

Since $90 < 180$, A works the full day.

Remaining Work: $180 - 90 = 90$ units

Total Days: $27 + 1 = 28$ days

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Day 29 (Cycle 10, Day 2): B works

Capacity: $10 \text{ hours} \times 6 \text{ units/hr} = 60$ units

Required: 90 units

Since $60 < 90$, B works the full day.

Remaining Work: $90 - 60 = 30$ units

Total Days: $28 + 1 = 29$ days

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Day 30 (Cycle 10, Day 3): C works

Capacity: $10 \text{ hours} \times 5 \text{ units/hr} = 50$ units

Required: 30 units

Since $50 > 30$, C will finish the work mid-day.

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The fractional day based on the rule: $\frac{q \text{ (hours worked)}}{p \text{ (scheduled hours)}}$.

Work Remaining: 30 units

C's Rate: 5 units/hr

Hours worked ($q$): $\frac{30}{5} = 6$ hours

Scheduled hours ($p$): 10 hours

$\text{Fractional Day} = \frac{6}{10}$

$= 0.6 \text{ days}$

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Total days = 29 full days + 0.6 fractional day

Total Days needed = 29.6