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In Episode 3 of our GMAT Ninja Critical Reasoning series, we tackle Discrepancy, Paradox, and Explain an Oddity questions. You know the feeling: the passage gives you two facts that seem completely contradictory....- May 08
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18 Mar 2025, 02:25
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
57% (02:33) correct
43%
(02:30)
wrong
based on 127
sessions
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1000 electrical engineering students are testing 1000 light bulbs. All bulbs, functional and connected to a power supply, are initially off. Student 1 flips the switches of bulbs 1, 2, 3, 4, and so on. Student 2 then flips bulbs 2, 4, 6, 8, etc. Student 3 flips bulbs 3, 6, 9, and so on, until all 1000 students have done this. What is the final status of bulbs 25, 93, and 576 after all students have finished flipping? (A single flip turns a light bulb from on to off, or off to on.)
A. off, off, off
B. on, off, on
C. off, off, on
D. on, on, on
E. on, on, off
A. off, off, off
B. on, off, on
C. off, off, on
D. on, on, on
E. on, on, off
18 Mar 2025, 03:43
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\(25 = 5^2\)
Adding 1 to the exponent:
\((2+1) = 3 = 1 \mod 2 \)
\(ON\).
---
\(93 = 3^1 \times 31^1\)
Adding 1 to the exponents:
\((1+1) \times (1+1) = 2 \times 2 = 4 = 0 \mod 2 \)
\(OFF\).
---
\(576 = 2^6 \times 3^2\)
Adding 1 to the exponents:
\((6+1) \times (2+1) = 7 \times 3 = 21 = 1 \mod 2 \)
\(ON\).
Adding 1 to the exponent:
\((2+1) = 3 = 1 \mod 2 \)
\(ON\).
---
\(93 = 3^1 \times 31^1\)
Adding 1 to the exponents:
\((1+1) \times (1+1) = 2 \times 2 = 4 = 0 \mod 2 \)
\(OFF\).
---
\(576 = 2^6 \times 3^2\)
Adding 1 to the exponents:
\((6+1) \times (2+1) = 7 \times 3 = 21 = 1 \mod 2 \)
\(ON\).
19 Mar 2025, 02:22
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Student 1 flips light bulb 1, 2, 3,...and so on => Bulbs that are multiple of 1
Student 2 flips light bulb 2, 4, 6,...and so on => Bulbs that are multiple of 2
Student 3 flips light bulb 3, 6, 9,...and so on => Bulbs that are multiple of 3
...
...
Student 1000 flips light bulb 1000 => Bulbs that are multiple of 1000
From above we can say that student k will flip the bulbs that are multiples of k => thus, a light bulb j will be flipped by all students for whom j is a multiple of that student number; in other words, light bulb j will be flipped by all student no. i, where i is a factor of j. Example: Bulb 6 will be flipped by students {1, 2, 3, 6} => factors of 6.
Now, light bulb is initially off, when one student flips it, it goes on and if a second student flips the same light bulb it goes off, thus if even number of flips occur => its final status is off, and if odd number of flips occur => its final status is on
Bulb no. 25 is flipped by Students {1, 5, 25} => odd number of flips, final status on
Bulb no. 93 is flipped by Students {1, 3, 31, 93} => even number of flips, final status off
Bulb no. 576 is flipped by \(576 = 2^6*3^2 = 7*3\) = 21 students => odd number of flips, final status on
Answer B.
Student 2 flips light bulb 2, 4, 6,...and so on => Bulbs that are multiple of 2
Student 3 flips light bulb 3, 6, 9,...and so on => Bulbs that are multiple of 3
...
...
Student 1000 flips light bulb 1000 => Bulbs that are multiple of 1000
From above we can say that student k will flip the bulbs that are multiples of k => thus, a light bulb j will be flipped by all students for whom j is a multiple of that student number; in other words, light bulb j will be flipped by all student no. i, where i is a factor of j. Example: Bulb 6 will be flipped by students {1, 2, 3, 6} => factors of 6.
Now, light bulb is initially off, when one student flips it, it goes on and if a second student flips the same light bulb it goes off, thus if even number of flips occur => its final status is off, and if odd number of flips occur => its final status is on
Bulb no. 25 is flipped by Students {1, 5, 25} => odd number of flips, final status on
Bulb no. 93 is flipped by Students {1, 3, 31, 93} => even number of flips, final status off
Bulb no. 576 is flipped by \(576 = 2^6*3^2 = 7*3\) = 21 students => odd number of flips, final status on
Answer B.
