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

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A lecture course consists of 595 students. The students are to be divi  [#permalink]

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19 Oct 2016, 05:04
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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

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Re: A lecture course consists of 595 students. The students are to be divi  [#permalink]

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19 Oct 2016, 05:14
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595=5*17*7

The answer choice should not be a factor of 595

45=9*5

C

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Re: A lecture course consists of 595 students. The students are to be divi  [#permalink]

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19 Oct 2016, 06:31
1
alanforde800Maximus 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

The question asks to find the number (among the given options) which will not divide 595

Only (C) 45 satisfies..

Hence answer will be (C) 45

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Re: A lecture course consists of 595 students. The students are to be divi  [#permalink]

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19 Oct 2016, 21:55
1
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.

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Re: A lecture course consists of 595 students. The students are to be divi  [#permalink]

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30 May 2018, 16:19
alanforde800Maximus 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

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.

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Re: A lecture course consists of 595 students. The students are to be divi  [#permalink]

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27 Sep 2018, 19:17
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...

Will 119 divide evenly into 595? Yes it will --> 5 times.

Will 85 divide evenly into 595? Yes it will --> 7 times.

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.

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Re: A lecture course consists of 595 students. The students are to be divi  [#permalink]

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31 Jan 2019, 05:24
alanforde800Maximus 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

$$\frac{{N\,\,{\text{students}}}}{{{\text{section}}}}$$

$$?\,\,:\,\,N\,\,{\text{impossible}}$$

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

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

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

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

The correct answer is therefore (C).

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

Regards,
Fabio.
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Re: A lecture course consists of 595 students. The students are to be divi   [#permalink] 31 Jan 2019, 05:24
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