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A lecture course consists of 595 students. The students are

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by BTGmoderatorDC » Sun Aug 11, 2019 3:28 am

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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

OA C

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by swerve » Sun Aug 11, 2019 12:47 pm
You can factorize \(595\) into prime factors \(5\cdot 7\cdot 17\)

The options are multiples of these prime numbers only, you will notice \(45\) is the only one which you can't get with these numbers.

Therefore, __C__
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by Scott@TargetTestPrep » Wed Aug 14, 2019 7:04 pm
BTGmoderatorDC 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

OA C

Source: Veritas Prep
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.

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 students in a discussion section, and so can 35 (which is 5 x 7) and 85 (which is 5 x 17). So, by process of elimination, it can't be 45.

Answer: C

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