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22 Oct 2024, 01:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
74% (01:14) correct
26%
(01:30)
wrong
based on 123
sessions
History
Date
Time
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Not Attempted Yet
If n is the product of the first 10 prime numbers, which of the following is an integer?
I. \(\frac{n}{7}\)
II. \(\frac{n}{4}\)
III. \(\frac{n}{143}\)
A. I only
B. I and II only
C. I and III only
D. III only
E. I, II, and III
I. \(\frac{n}{7}\)
II. \(\frac{n}{4}\)
III. \(\frac{n}{143}\)
A. I only
B. I and II only
C. I and III only
D. III only
E. I, II, and III
This is a PS Butler Question
Check the links to other Butler Projects:
Check the links to other Butler Projects:
- Data Sufficiency Butler
- Critical Reasoning Butler
- Reading Comprehension Butler
- Integrated Reasoning Butler
22 Oct 2024, 02:01
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n is the product of the first 10 prime numbers:
n = 2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23 * 29
I. n/7 is an integer since 7 is in the product of n
II. n/4 is NOT an integer since we need 2^2 in the product of n
III. n/143 is an integer since the prime factorization of 143 is 11 * 13
I and III only
Option C
n = 2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23 * 29
I. n/7 is an integer since 7 is in the product of n
II. n/4 is NOT an integer since we need 2^2 in the product of n
III. n/143 is an integer since the prime factorization of 143 is 11 * 13
I and III only
Option C
22 Oct 2024, 02:04
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n = 2.3.5.7.11.13....
1. n has 7 so 1 is true
2. n has only one 2 as the only even prime number. There wont be any other 2 or a multiple of 2 in the remaining prime numbers. so, 2 is incorrect
3. n has 11 and 13 whose product is 143. So, 3 is true
hence, 1 and 3 only
1. n has 7 so 1 is true
2. n has only one 2 as the only even prime number. There wont be any other 2 or a multiple of 2 in the remaining prime numbers. so, 2 is incorrect
3. n has 11 and 13 whose product is 143. So, 3 is true
hence, 1 and 3 only
