Daily Math Puzzle: 2026-09-29
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2026-09-29
Detective Reese is on a crucial case, analyzing security footage from a suspect's secret lab. The lab has three security cameras: Alpha, Beta, and Gamma, all feeding into a single, fixed-capacity server. Camera Alpha records at a base rate 'R' GB/hour. Camera Beta records at twice Alpha's rate (2R GB/hour), and Camera Gamma records at thrice Alpha's rate (3R GB/hour). Reese discovered the server was completely full. Examining the server logs, he pieced together the sequence of events: 1. Initially, only Camera Alpha was active, recording for a specific duration (let's call it 'T' hours). 2. Then, Camera Beta was activated, joining Alpha. Both Alpha and Beta recorded together for *half* the time Alpha had recorded alone (T/2 hours). 3. Finally, Camera Gamma was activated, joining Alpha and Beta. All three cameras recorded together until the server reached its full capacity. Crucially, the server's logs also revealed a unique insight: the total amount of data recorded by Camera Alpha throughout *all phases* of its operation was exactly equal to the total amount of data recorded by Camera Beta throughout *all phases* of its operation. What proportion of the server's total storage capacity was ultimately filled *only* by Camera Gamma's footage?
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Solution
3/11 — Let 'R' be the recording rate of Camera Alpha in GB/hour. Based on the problem:
- Camera Alpha's rate = R
- Camera Beta's rate = 2R
- Camera Gamma's rate = 3R
Let 'T' be the duration (in hours) Camera Alpha recorded alone in Phase 1.
Let 'X' be the duration (in hours) that all three cameras recorded together in Phase 3.
**Step 1: Calculate the total data recorded by Alpha and Beta throughout their active periods.**
**Camera Alpha's total data contribution:**
- In Phase 1 (Alpha alone): R * T
- In Phase 2 (Alpha + Beta): R * (T/2)
- In Phase 3 (Alpha + Beta + Gamma): R * X
Total Alpha data = RT + (RT/2) + RX = 1.5RT + RX
**Camera Beta's total data contribution:**
- In Phase 1 (Beta not active): 0
- In Phase 2 (Alpha + Beta): 2R * (T/2) = RT
- In Phase 3 (Alpha + Beta + Gamma): 2R * X
Total Beta data = RT + 2RX
**Step 2: Use the crucial log revelation to find the relationship between T and X.**
The problem states: Total Alpha data = Total Beta data
1.5RT + RX = RT + 2RX
Subtract RT from both sides: 0.5RT + RX = 2RX
Subtract RX from both sides: 0.5RT = RX
Since R is a rate (and thus > 0), we can divide by R: 0.5T = X
This means the duration of Phase 3 (X) is half the duration of Phase 1 (T/2).
**Step 3: Calculate the total data recorded in each phase and the server's total capacity.**
**Phase 1 (Alpha alone):**
- Duration: T
- Combined rate: R
- Data (D1) = R * T = RT
**Phase 2 (Alpha + Beta):**
- Duration: T/2
- Combined rate: R + 2R = 3R
- Data (D2) = 3R * (T/2) = 1.5RT
**Phase 3 (Alpha + Beta + Gamma):**
- Duration: X = T/2 (from Step 2)
- Combined rate: R + 2R + 3R = 6R
- Data (D3) = 6R * (T/2) = 3RT
**Total Server Capacity (C):**
C = D1 + D2 + D3 = RT + 1.5RT + 3RT = 5.5RT
**Step 4: Calculate Camera Gamma's total footage contribution.**
Camera Gamma was only active during Phase 3.
- Gamma's rate in Phase 3 = 3R
- Gamma's duration in Phase 3 = X = T/2
Gamma's footage = 3R * (T/2) = 1.5RT
**Step 5: Determine the proportion.**
The proportion of the server's total storage capacity filled by Gamma's footage is:
(Gamma's footage) / (Total Capacity) = (1.5RT) / (5.5RT)
Divide both numerator and denominator by RT:
= 1.5 / 5.5
Multiply by 10 to remove decimals:
= 15 / 55
Simplify the fraction by dividing by 5:
= 3 / 11
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