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# If x and y are positive integers and x/y has a remainder of 5, what is

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If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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Updated on: 28 Jan 2019, 00:02
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If x and y are positive integers and x/y has a remainder of 5, what is the smallest possible value of xy?

A. 6
B. 12
C. 18
D. 24
E. 30

This ques is from Man prep book. I don't understand why smallest value of x is 5 and not 11.

Originally posted by joyseychow on 20 Jan 2010, 20:54.
Last edited by Bunuel on 28 Jan 2019, 00:02, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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01 Feb 2010, 05:17
7
3
$$x=qy+5$$, not $$y+5$$, and note that $$q$$ can be zero. In this case $$x=0*y+5=5$$, this is smallest value of $$x$$.

Hey Bunuel....you lose me after the last part above... I still don't see how y=6. Could you explain step by step? or does anyone have a solution that doesn't involve algebra?

$$\frac{5}{y}$$ gives remainder 5 means $$y>x$$. For EVERY $$y$$ more than $$x=5$$, $$\frac{5}{y}$$ will give remainder of $$5$$: $$\frac{5}{6}$$, $$\frac{5}{7}$$, $$\frac{5}{16778}$$, ... all these fractions have remainder of $$5$$.

I thought $$\frac{5}{6}$$, is equal to .8 with a remainder of 2?

THEORY:
If $$a$$ and $$d$$ are positive integers, there exists unique integers $$q$$ and $$r$$, such that $$a = qd + r$$ and $$0\leq{r}<d$$. Where $$a$$ is a dividend, $$d$$ is a divisor, $$q$$ is a quotient ($$\geq0$$) and $$r$$ is a remainder.

Now consider the case when dividend $$a$$ is less than divisor $$d$$. For instance: what is the ramainder when $$a=7$$ divided by $$d=9$$:

$$a = qd + r$$ --> $$7=q*9+r$$, as $$q\geq0$$ and $$0\leq{r}<d$$ --> $$7=0*9+r=0*9+7$$ --> $$r=remainder=7$$.

Hence in ANY case when positive integer $$a$$ is divided by positive integer $$d$$ and given that $$a<d$$, remainder will always be $$a$$. Note here that EVERY GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

So: $$\frac{7}{9}$$, remainder $$7$$; $$\frac{11}{13}$$, remainder $$11$$; $$\frac{135}{136}$$, remainder $$135$$.

Back to our original question:
If x and y are positive integers and x/y has a remainder of 5, what is the smallest possible value of xy?

A. 6
B. 12
C. 18
D. 24
E. 30

When we got that $$x=5$$ and $$\frac{5}{y}$$ gives remainder of $$5$$, according to the above we can conclude that $$y$$ must be more than $$x=5$$. As we want to minimize the $$x*y$$ we need least integer $$y$$ which is more than $$x=5$$. Thus $$y=6$$ and $$xy=30$$.

Check it:
$$5=0*6+5$$ --> $$x=5$$ dividend, $$y=6$$ divisor, $$q=0$$ quotient and $$r=5$$ remainder.

Hope it helps.
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Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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21 Jan 2010, 01:49
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joyseychow wrote:
If x and y are positive integers and x/y has a remainder of 5, what is the smallest possible value of xy?

[spoiler]This ques is from Man prep book. I don't understand why smallest value of x is 5 and not 11.[/spoiler]

Given $$x=qy+5$$, where $$q$$ is a quotient, an integer $$\geq0$$. Which means that the least value of $$x$$ is when $$q=0$$, in that case $$x=5$$. This basically means that numerator $$x$$, is less than denominator $$y$$.

Now the smallest denominator $$y$$, which is more than numerator $$x=5$$ is $$6$$. so we have $$x=5$$ and $$y=6$$ --> $$xy=30$$.
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Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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30 Jan 2010, 15:45
Bunuel wrote:
Given $$x=qy+5$$, where $$q$$ is a quotient, an integer $$\geq0$$. Which means that the least value of $$x$$ is when $$q=0$$, in that case $$x=5$$. This basically means that numerator $$x$$, is less than denominator $$y$$.

Now the smallest denominator $$y$$, which is more than numerator $$x=5$$ is $$6$$. so we have $$x=5$$ and $$y=6$$ --> $$xy=30$$.

How do you manage to do such things???? What is the best approach to master such things??? Please guide!
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Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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30 Jan 2010, 15:59
1
1
jeeteshsingh wrote:
Bunuel wrote:
Given $$x=qy+5$$, where $$q$$ is a quotient, an integer $$\geq0$$. Which means that the least value of $$x$$ is when $$q=0$$, in that case $$x=5$$. This basically means that numerator $$x$$, is less than denominator $$y$$.

Now the smallest denominator $$y$$, which is more than numerator $$x=5$$ is $$6$$. so we have $$x=5$$ and $$y=6$$ --> $$xy=30$$.

How do you manage to do such things???? What is the best approach to master such things??? Please guide!

What "things" exactly do you mean?

This question is about the remainders, so would suggest to review this concept in guids and/or to practice as much as possible these types of qestions (refer to the tags list by forum: REMAINDERS). Also you can check sriharimurthy's topic about the remainders at: compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html

Hope it helps.
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Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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31 Jan 2010, 08:26
Bunuel wrote:

Given $$x=qy+5$$, where $$q$$ is a quotient, an integer $$\geq0$$. Which means that the least value of $$x$$ is when $$q=0$$, in that case $$x=5$$. This basically means that numerator $$x$$, is less than denominator $$y$$.

Now the smallest denominator $$y$$, which is more than numerator $$x=5$$ is $$6$$. so we have $$x=5$$ and $$y=6$$ --> $$xy=30$$.

Bunuel... am again lost here...

As said x = qy + 5...... and since $$q\geq0$$... this should be mean that X = Y + 5... then how come X can be less than Y.... we are adding 5 to Y to get X...
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Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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31 Jan 2010, 08:43
3
jeeteshsingh wrote:
Bunuel wrote:

Given $$x=qy+5$$, where $$q$$ is a quotient, an integer $$\geq0$$. Which means that the least value of $$x$$ is when $$q=0$$, in that case $$x=5$$. This basically means that numerator $$x$$, is less than denominator $$y$$.

Now the smallest denominator $$y$$, which is more than numerator $$x=5$$ is $$6$$. so we have $$x=5$$ and $$y=6$$ --> $$xy=30$$.

Bunuel... am again lost here...

As said x = qy + 5...... and since $$q\geq0$$... this should be mean that X = Y + 5... then how come X can be less than Y.... we are adding 5 to Y to get X...

$$x=qy+5$$, not $$y+5$$, and note that $$q$$ can be zero. In this case $$x=0*y+5=5$$, this is smallest value of $$x$$.

$$\frac{5}{y}$$ gives remainder 5 means $$y>x$$. For EVERY $$y$$ more than $$x=5$$, $$\frac{5}{y}$$ will give remainder of $$5$$: $$\frac{5}{6}$$, $$\frac{5}{7}$$, $$\frac{5}{16778}$$, ... all these fractions have remainder of $$5$$.

Hope it's clear.
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Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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31 Jan 2010, 20:02
Bunuel wrote:
jeeteshsingh wrote:
Bunuel wrote:

Given $$x=qy+5$$, where $$q$$ is a quotient, an integer $$\geq0$$. Which means that the least value of $$x$$ is when $$q=0$$, in that case $$x=5$$. This basically means that numerator $$x$$, is less than denominator $$y$$.

Now the smallest denominator $$y$$, which is more than numerator $$x=5$$ is $$6$$. so we have $$x=5$$ and $$y=6$$ --> $$xy=30$$.

Bunuel... am again lost here...

As said x = qy + 5...... and since $$q\geq0$$... this should be mean that X = Y + 5... then how come X can be less than Y.... we are adding 5 to Y to get X...

$$x=qy+5$$, not $$y+5$$, and note that $$q$$ can be zero. In this case $$x=0*y+5=5$$, this is smallest value of $$x$$.

Hey Bunuel....you lose me after the last part above... I still don't see how y=6. Could you explain step by step? or does anyone have a solution that doesn't involve algebra?

$$\frac{5}{y}$$ gives remainder 5 means $$y>x$$. For EVERY $$y$$ more than $$x=5$$, $$\frac{5}{y}$$ will give remainder of $$5$$: $$\frac{5}{6}$$, $$\frac{5}{7}$$, $$\frac{5}{16778}$$, ... all these fractions have remainder of $$5$$.

I thought $$\frac{5}{6}$$, is equal to .8 with a remainder of 2?
Manager
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Posts: 129
Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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12 Feb 2010, 18:36
1
If x and y are positive integers and x/y has a remainder of 5, what is the smallest possible value of xy?

y>5 and smallest number greater than 5 is 6.

x=y(quotient)+reminder = 6(quotient)+5
to get smallest number, substitute 0
=>x=6(0)+5=5

and xy=5x6=30
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Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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02 Aug 2012, 17:45
1
Can someone help me understand this?

The question is if x and y are positive integers and x/y has a remainder of 5, what is the smallest possible of xy?

The answer is 5*6 = 30

I came up with 66 because I understand that y has to be at least 6 and with that I would think that y has to be 11 which is how I derived 66. The answer says that the smallest y is 5 but I don't understand how that is possible.
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Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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02 Aug 2012, 19:17
GmatSlayer112 wrote:
Can someone help me understand this?

The question is if x and y are positive integers and x/y has a remainder of 5, what is the smallest possible of xy?

The answer is 5*6 = 30

I came up with 66 because I understand that y has to be at least 6 and with that I would think that y has to be 11 which is how I derived 66. The answer says that the smallest y is 5 but I don't understand how that is possible.

if y is 5, then 5/6 = 0 r 5 -- that is, there are NO complete sixes in the number five, so all five are left over in the remainder!
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Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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02 Aug 2012, 23:08
1
GmatSlayer112 wrote:
Can someone help me understand this?

The question is if x and y are positive integers and x/y has a remainder of 5, what is the smallest possible of xy?

The answer is 5*6 = 30

I came up with 66 because I understand that y has to be at least 6 and with that I would think that y has to be 11 which is how I derived 66. The answer says that the smallest y is 5 but I don't understand how that is possible.

When dividing a positive integer $$x$$ by another positive integer $$y$$, we get a unique non-negative integer quotient $$q$$ and a unique remainder $$r$$, where $$r$$ is an integer such that $$0\leq{r}<y$$. When $$r = 0$$, we say that $$x$$ is divisible by $$y$$, that $$y$$ is a factor of $$x$$, or that $$x$$ is a multiple of $$y$$. In this case, $$x$$ can be evenly divided by $$y$$.

In our case, we can write $$x = q*y + 5$$, and as the remainder is always smaller than the divisor, we can deduce that $$y$$ should be at least 6. The minimum quotient $$q$$ is 0 which gives $$x = 5$$. Therefore, the smallest possible value of $$xy$$ is $$5 * 6 =30$$.
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Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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02 Aug 2012, 23:32
3
1
GmatSlayer112 wrote:
Can someone help me understand this?

The question is if x and y are positive integers and x/y has a remainder of 5, what is the smallest possible of xy?

The answer is 5*6 = 30

I came up with 66 because I understand that y has to be at least 6 and with that I would think that y has to be 11 which is how I derived 66. The answer says that the smallest y is 5 but I don't understand how that is possible.

Think: What is the quotient and what is the remainder when 3 is divided by 10?
The quotient is 0 (an integer) and the remainder is 3.

What is the quotient and what is the remainder when 5 is divided by 12?
The quotient is 0 (an integer) and the remainder is 5.

When a smaller number is divided by a larger number, the quotient is 0 and the remainder is the smaller number.

Divisibility is basically grouping. When you divide n by 10, you make as many groups of 10 as you can and the leftover is the remainder. When you have 3 and you want to divide it by 10, you make 0 groups of 10 each and you have 3 leftover which is the remainder.
See this post: http://www.veritasprep.com/blog/2011/04 ... unraveled/
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Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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04 Aug 2012, 18:33
Bunuel wrote:
joyseychow wrote:
If x and y are positive integers and x/y has a remainder of 5, what is the smallest possible value of xy?

[spoiler]This ques is from Man prep book. I don't understand why smallest value of x is 5 and not 11.[/spoiler]

Given $$x=qy+5$$, where $$q$$ is a quotient, an integer $$\geq0$$. Which means that the least value of $$x$$ is when $$q=0$$, in that case $$x=5$$. This basically means that numerator $$x$$, is less than denominator $$y$$.

Now the smallest denominator $$y$$, which is more than numerator $$x=5$$ is $$6$$. so we have $$x=5$$ and $$y=6$$ --> $$xy=30$$.

for this first time, i exactly though the same way as Bunuel...good teacher bunuel...thanks for this,,,,
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Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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12 Jan 2014, 05:11
Please help me understand one point, the question asks- min. possible value of xy and given x when divided by y gives remainder 5 so

x=my+5;
then suppose y=2 and m=1 and then x=1*2+5 =7;
xy=7*2=14 which is less than 30 then mentioned answer in the post, please help me explain where I am doing the mistake in calculating the min possible value for xy.

Thanks and Regards
Kiran Saxena
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Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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12 Jan 2014, 06:06
kiransaxena1988 wrote:
Please help me understand one point, the question asks- min. possible value of xy and given x when divided by y gives remainder 5 so

x=my+5;
then suppose y=2 and m=1 and then x=1*2+5 =7;
xy=7*2=14 which is less than 30 then mentioned answer in the post, please help me explain where I am doing the mistake in calculating the min possible value for xy.

Thanks and Regards
Kiran Saxena

The point is that if x=7 and y=2, then the remainder when x=7 is divided y=2 is 1 not 5, thus your example is not valid.

Hope it's clear.
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Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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26 Sep 2016, 04:18
joyseychow wrote:
If x and y are positive integers and x/y has a remainder of 5, what is the smallest possible value of xy?

This ques is from Man prep book. I don't understand why smallest value of x is 5 and not 11.

This is a classic trap: 0 is a multiple of every natural number. 5 can be written as (0*6)+5 which also means that when 5 is divided by 6, remainder is 5.
In order to find the minimum possible value of xy, we will try to make both x and y the smallest. To leave a remainder of 5, minimum possible value of y will be 6. Smallest possible number that will result in the remainder of 5 is then 5 (as explained above). So, xy=5*6=30

Hope it helps
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Schools: EDHEC (A$) GMAT 1: 650 Q43 V37 GPA: 3.2 WE: General Management (Human Resources) Re: If x and y are positive integers and x/y has a remainder of 5, what is [#permalink] ### Show Tags 26 Jan 2018, 08:10 Can someone help me? My reasoning was that: 1/5 = 0 (5) so x=1 and y=5 and so the minimum value of xy=5! Why I am wrong? Bunuel Math Expert Joined: 02 Sep 2009 Posts: 59712 Re: If x and y are positive integers and x/y has a remainder of 5, what is [#permalink] ### Show Tags 26 Jan 2018, 08:16 1 MvArrow wrote: Can someone help me? My reasoning was that: 1/5 = 0 (5) so x=1 and y=5 and so the minimum value of xy=5! Why I am wrong? Bunuel 1 divided by 5 gives the remainder of 1, not 5. _________________ Current Student Joined: 06 Sep 2016 Posts: 125 Location: Italy Schools: EDHEC (A$)
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Re: If x and y are positive integers and x/y has a remainder of 5, what is  [#permalink]

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26 Jan 2018, 08:54
Bunuel wrote:
MvArrow wrote:
Can someone help me?
My reasoning was that:
1/5 = 0 (5)
so x=1 and y=5 and so the minimum value of xy=5!

Why I am wrong? Bunuel

1 divided by 5 gives the remainder of 1, not 5.

True! Today I have my mind in the clouds
Re: If x and y are positive integers and x/y has a remainder of 5, what is   [#permalink] 26 Jan 2018, 08:54

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