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18 Sep 2025, 07:20
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2 iterations: 5,836
4 iterations: 5,324
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
65% (03:16) correct
35%
(03:28)
wrong
based on 260
sessions
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Date
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Not Attempted Yet
Umbrella Corporation monitors a network of 6,000 computers. At the start of the virus outbreak, 4 of these computers get infected. The infection weakens with each round of spreading:
Select for 2 iterations the number of computers that remain uninfected after two iterations are completed, and select for 4 iterations the number of computers that remain uninfected after four iterations are completed. Make only two selections, one in each column.
- In the first iteration, each of the 4 initially infected machines infects 8 others.
- In the second iteration, each machine infected in the first iteration infects 4 others.
- In the third iteration, each machine infected in the second iteration infects 2 others.
- In the fourth iteration, each machine infected in the third iteration infects 1 other.
Select for 2 iterations the number of computers that remain uninfected after two iterations are completed, and select for 4 iterations the number of computers that remain uninfected after four iterations are completed. Make only two selections, one in each column.
| 2 iterations | 4 iterations | |
| 5,260 | ||
| 5,324 | ||
| 5,452 | ||
| 5,488 | ||
| 5,836 | ||
| 5,872 |
ShowHide Answer
Official Answer
2 iterations: 5,836
4 iterations: 5,324
26 Sep 2025, 02:15
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Bunuel
The network contains 6,000 computers in total, and the outbreak begins with 4 of them infected.
Iteration 1: Each of the 4 infects 8 others, producing 32 new infections. Adding these to the initial 4 gives 36 infected computers.
Iteration 2: The 32 from the first round each infect 4 others, producing 128 new infections. Adding this to the 36 infected so far gives a cumulative total of 164 after two iterations. The number of unaffected computers is 6,000 - 164 = 5,836.
Iteration 3: The 128 from the second round each infect 2 others, producing 256 new infections. Cumulative total = 164 + 256 = 420.
Iteration 4: The 256 from the third round each infect 1 other, producing 256 new infections. Cumulative total = 420 + 256 = 676 after four iterations. The number of unaffected computers is 6,000 - 676 = 5,324.
I29-166
General Discussion
20 Sep 2025, 02:32
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Bunuel
Total computers within the network = 6000
At the START, the number of infections = 4
First iteration :
Each of the 4 initially infected machine infects 8 others.
So, the number of infections = 4*8 = 32.
Second Iteration:
The number of infected computers at the beginning of second iteration = 32.
The number of Infections at second iteration = 32*4 = 128.
The number of infected machines TILL the Second iteration = 4+32+128 = 164.
The number of UNINFECTED machine = 6000 - 164 = 5836.
Third Iteration:
The number of infected machines at the beginning of third iteration = 128.
The number of infections at the third iteration = 128*2 = 256.
The number of infected machines TILL the Third iteration = 4+32+128+256= 420.
The number of UNINFECTED machine = 6000 - 420 = 5580.
Fourth Iteration:
The number of infected machines at the beginning of the fourth iteration = 256.
The number of infections at the fourth iteration = 256*1 = 256.
The number of infected machines TILL the end of Fourth Iteration = 4+32+128+256+256 = 676.
The number of UNINFECTED machine = 6000 - 676 = 5324.
The number of UNINFECTED machines after 2nd Iteration = 5836.
The number of UNINFECTED machines after 4th Iteration = 5324.
