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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
(students per section)(number of sections) = 595 students
number of sections = 595/(students per section).
The equation in blue indicates that 595 must be divisible by the number of students per section.

An integer is divisible by 3 only if its digit sum is a multiple of 3.
The digit sum of 595 = 5+9+5 = 19.
Since the digit sum of 595 is NOT a multiple of 3, 595 is not divisible by any integer that is a multiple of 3.
45 = 3*3*5 = multiple of 3.
Thus, 595 is not divisible by 45, implying that the number of students per section cannot be 45.

The correct answer is C.

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Hi All,

We're told that a lecture course consists of 595 students and that 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 is really just about 'multiples' - and if you understand division rules and/or prime factorization, then you can potentially answer this question quickly. That having been said, sometimes the fastest way to get to the correct answer is to use simple 'brute force' Arithmetic and do just enough work to PROVE which answer is correct...

Answer E: 119
Will 119 divide evenly into 595? Yes it will --> 5 times.
Eliminate Answer E.

Answer D: 85
Will 85 divide evenly into 595? Yes it will --> 7 times.
Eliminate Answer D.

Answer C: 45
Will 45 divide evenly into 595? NO it will NOT (there's a remainder) --> 13r10
This is the answer that CANNOT be the number of students.

Final Answer: C


GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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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
\[\frac{{N\,\,{\text{students}}}}{{{\text{section}}}}\]
\[?\,\,:\,\,N\,\,{\text{impossible}}\]

\[{\text{595}}\,\,{\text{students}}\,\,\,\,{\text{ = }}\,\,\,\,{\text{M}}\,\,{\text{sections}}\,\,\left( {\frac{{N\,\,{\text{students}}}}{{{\text{section}}}}} \right)\]
\[\left\{ \begin{gathered}
M \cdot N = 595 \hfill \\
M,N\,\, \geqslant 1\,\,{\text{ints}} \hfill \\
\end{gathered} \right.\,\,\,\,\, \Rightarrow \,\,\,\,M\,\,{\text{and}}\,\,N\,\,{\text{are}}\,\,{\text{pairs}}\,\,{\text{of}}\,\,\underline {{\text{divisors}}} \,\,{\text{of}}\,\,595\]
\[\frac{{500 + 50 + 45}}{5} = 119\,\,\,\, \Rightarrow \,\,\,\,595 = 5 \cdot 119\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{refute}}\,\,\left( E \right)\]
\[\frac{{119}}{7} = \frac{{70 + 28 + 21}}{7} = 17\,\,\,\, \Rightarrow \,\,\,\,119 = 7 \cdot 17\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{refute}}\,\,\left( A \right),\left( B \right),\left( D \right)\,\,\,\,\,\,\left[ {\frac{{85}}{5} = \frac{{50 + 35}}{5} = 17 \Rightarrow 85 = 5 \cdot 17} \right]\]

The correct answer is therefore (C).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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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
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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scott@targettestprep.com



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