A lecture course consists of 595 students. The students are to be

You can factorize 595 into prime factors 5*7*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.

HTH

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 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

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 E: 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:

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

\(\frac{{N\,\,{\text{students}}}}{{{\text{section}}}}\)

\(?\,\,:\,\,N\,\,{\text{impossible}}\)


\({\text{595}}\,\,{\text{students}}\,\,\,\,{\text{ = }}\,\,\,\,{\text{M}}\,\,{\text{sections}}\,\,\left( {\frac{{N\,\,{\text{students}}}}{{{\text{section}}}}} \right)\)

\(\left\{ \begin{gathered}\\
M \cdot N = 595 \hfill \\\\
M,N\,\, \geqslant 1\,\,{\text{ints}} \hfill \\ \\
\end{gathered} \right.\,\,\,\,\, \Rightarrow \,\,\,\,M\,\,{\text{and}}\,\,N\,\,{\text{are}}\,\,{\text{pairs}}\,\,{\text{of}}\,\,\underline {{\text{divisors}}} \,\,{\text{of}}\,\,595\)


\(\frac{{500 + 50 + 45}}{5} = 119\,\,\,\, \Rightarrow \,\,\,\,595 = 5 \cdot 119\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{refute}}\,\,\left( E \right)\)

\(\frac{{119}}{7} = \frac{{70 + 28 + 21}}{7} = 17\,\,\,\, \Rightarrow \,\,\,\,119 = 7 \cdot 17\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{refute}}\,\,\left( A \right),\left( B \right),\left( D \right)\,\,\,\,\,\,\left[ {\frac{{85}}{5} = \frac{{50 + 35}}{5} = 17 \Rightarrow 85 = 5 \cdot 17} \right]\)


The correct answer is therefore (C).


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

Regards,
Fabio.

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

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.