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A lecture course consists of 595 students. The students are

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A lecture course consists of 595 students. The students are

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Source: Veritas Prep

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

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BTGmoderatorLU wrote:
Source: Veritas Prep

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
In order to have an EQUAL number of students in each section, the number of students per section MUST BE A FACTOR of 595

Let's do some prime factorization
595 = (5)(7)(17)
From the prime factorization, we can see that answer choice A, B, D and E are all factors of 595

Answer: C

Cheers,
Brent

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Post
BTGmoderatorLU wrote:
Source: Veritas Prep

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
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

_________________

Scott Woodbury-Stewart
Founder and CEO
scott@targettestprep.com



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