A lecture course consists of 595 students. The students are to be divided into discussion sections, each with an equal number of students. Which of the following cannot be the number of students in a discussion section?
A. 17
B. 35
C. 45
D. 85
E. 119
The OA is C.
Please, can anyone assist with above problem? Thanks in advance.
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A lecture course consists of 595 students. The students are
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BTGmoderatorLU
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Number of discussion sections = (total number of students)/(number of students per section).BTGmoderatorLU wrote:A lecture course consists of 595 students. The students are to be divided into discussion sections, each with an equal number of students. Which of the following cannot be the number of students in a discussion section?
A. 17
B. 35
C. 45
D. 85
E. 119
section.
Since the total number of students = 595, the number of students per section must be a FACTOR OF 595.
The question stem asks for a value that CANNOT be the number of students per classroom.
For an integer to be a multiple of 3, its digits must sum to a multiple of 3.
Since the sum of the digits of 595 is not a multiple of 3, 595 is not divisible by 3.
Implication:
Any answer choice that is a multiple of 3 cannot be a factor of 595 and thus cannot serve as the number of students per section.
Since 45 is a multiple of 3, it cannot serve as the number of students per section.
The correct answer is C. $$$$
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Hi All,
We're told that a lecture course consists of 595 students and the students are to be divided into discussion sections, each with an EQUAL number of students. We're asked which of the following CANNOT be the number of students in a discussion section. This question can be solved in a couple of different ways, including with Prime Factorization.
One way to determine all of the numbers that divide evenly into 595 is to prime-factor the total and list out the possible products from the factors:
595 =
(5)(119) =
(5)(7)(17)
We can now determine the list of factors of 595:
1, 5, 7, 17
(5)(7) = 35
(5)(17) = 85
(7)(17) = 119
595
You'll notice 4 of the answers are in the list and one is NOT...
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
We're told that a lecture course consists of 595 students and the students are to be divided into discussion sections, each with an EQUAL number of students. We're asked which of the following CANNOT be the number of students in a discussion section. This question can be solved in a couple of different ways, including with Prime Factorization.
One way to determine all of the numbers that divide evenly into 595 is to prime-factor the total and list out the possible products from the factors:
595 =
(5)(119) =
(5)(7)(17)
We can now determine the list of factors of 595:
1, 5, 7, 17
(5)(7) = 35
(5)(17) = 85
(7)(17) = 119
595
You'll notice 4 of the answers are in the list and one is NOT...
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
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Jeff@TargetTestPrep
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Any number whose digits sum to a number divisible by 3 is itself divisible by 3. For example, 3,912 is divisible by 3 because the sum of its digits is 3 + 9 + 1 + 2 = 15, which is divisible by 3.BTGmoderatorLU wrote:A lecture course consists of 595 students. The students are to be divided into discussion sections, each with an equal number of students. Which of the following cannot be the number of students in a discussion section?
A. 17
B. 35
C. 45
D. 85
E. 119
Since 5 + 9 + 5 = 19, we see that 595 is not a multiple of 3. Since 45 is a multiple of 3, we cannot have 45 students in a discussion section.
Alternate solution:
Since 595 = 5 x 119 = 5 x 7 x 17, we see that 17 and 119 obviously can be the number of of students in discussion section and so can 35 (which is 5 x 7) and 85 (which is 5 x 17). So the only one it can't be is 45.
Answer: C
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