Which of the following CANNOT result in an integer?

Hello!

How did you understand that one prime number is dividing another prime number?

I am a bit confused with the wording of the question.

Can´t it be like the following?

2*3 divided by 6 = quotient would be 1.

Regards!

The wording of the question is not perfect.

It should read something like this: The result when one prime number is divided by a different prime number.

One prime number will never be divisible by a different prime number because a prime is divisible by only two numbers - 1 and itself.
So the result will never be an integer.

e.g. 3/2 or 5/7 or 13/5 etc

Note that 2*3 is not a prime number.

Also, quotient of a division is always an integer. The remainder is also an integer. Together, you get the result of division which may or may not be an integer.

So 13/5 gives quotient 2 and remainder 3.
But 13/5 = 2.6 (not an integer)
Hence, the use of the term quotient is also not right.

[/quote]The wording of the question is not perfect.

It should read something like this: The result when one prime number is divided by a different prime number.

One prime number will never be divisible by a different prime number because a prime is divisible by only two numbers - 1 and itself.
So the result will never be an integer.

e.g. 3/2 or 5/7 or 13/5 etc

Note that 2*3 is not a prime number.

Also, a quotient of a division is always an integer. The remainder is also an integer. Together, you get the result of division which may or may not be an integer.

So 13/5 gives quotient 2 and remainder 3.
But 13/5 = 2.6 (not an integer)
Hence, the use of the term quotient is also not right.[/quote]

Thank you! Now everything is very clear .

\(?\,\,:\,\,\,{\rm{cannot}}\,\,{\rm{be}}\,\,{\rm{integer}}\)

\(\left( {\rm{A}} \right)\,\,\,\left( {0 \cdot 0} \right) \div {1^{ - 1}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{A}} \right)\,\,\,{\rm{refuted}}\)

\(\left( {\rm{B}} \right)\,\,\,0 \div 7\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{B}} \right)\,\,\,{\rm{refuted}}\)

\(\left( {\rm{D}} \right)\,\,0 \div 3\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{D}} \right)\,\,\,{\rm{refuted}}\)

\(\left( {\rm{E}} \right)\,\,\left( {1 + 1} \right) \div 2\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{E}} \right)\,\,\,{\rm{refuted}}\)

Conclusion: (C) is the corrrect answer by exclusion.

POST-MORTEM:

\(\left( {\rm{C}} \right)\,\,\,{{{p_1}} \over {{p_2}}} \ne {\mathop{\rm int}} \,\,\,\,:\,\,\,\,\,{\rm{if}}\,\,\,\,{{{p_1}} \over {{p_2}}} = {\mathop{\rm int}} \,\,\,\,\left\{ \matrix{\\
{p_1}\,\,,{p_2}\,\, > 0\,\,\,\, \Rightarrow \,\,\,{\mathop{\rm int}} \,\, \ge 1\,\,\,\,\left( * \right) \hfill \cr \\
{p_1} \ne {p_2}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,{\mathop{\rm int}} \,\, \ge 2 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{p_1} = {\mathop{\rm int}} \,\, \cdot {p_2}\,\,\,\,\,\,\,\,\)

\(\,\mathop \Rightarrow \limits_{{p_2}\,\, \ge \,\,2}^{{\mathop{\rm int}} \,\, \ge \,\,2} \,\,\,\,\,\,\,\,{p_1}\,\,{\rm{not}}\,\,{\rm{prime}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{impossible}}\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Recall that a prime number is a number that has only two factors: 1 and itself. Therefore, the quotient of two distinct prime numbers, such as 2/7 or 13/11, can’t be an integer since the two prime numbers can’t be multiples of each other.

Answer: C

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.