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

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

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This ques is from Man prep book. I don't understand why smallest value of x is 5 and not 11.
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E
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Bunuel
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If x and y are positive integers and x/y has a remainder of 5, what is 01 Feb 2010, 05:17
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CoIIegeGrad09

\(x=qy+5\), 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?
Show more

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.

Answer: E.

Hope it helps.
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If x and y are positive integers and x/y has a remainder of 5, what is 21 Jan 2010, 01:49
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joyseychow
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]
Show more


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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If x and y are positive integers and x/y has a remainder of 5, what is 30 Jan 2010, 15:45
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Bunuel

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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How do you manage to do such things???? :shock: :roll: What is the best approach to master such things??? Please guide!
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If x and y are positive integers and x/y has a remainder of 5, what is 30 Jan 2010, 15:59
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jeeteshsingh
Bunuel

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???? :shock: :roll: What is the best approach to master such things??? Please guide!
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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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If x and y are positive integers and x/y has a remainder of 5, what is 31 Jan 2010, 08:26
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Bunuel



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 \(y=6\) --> \(xy=30\).
Show more

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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If x and y are positive integers and x/y has a remainder of 5, what is 31 Jan 2010, 08:43
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jeeteshsingh
Bunuel



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... :?
Show more

\(x=qy+5\), 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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If x and y are positive integers and x/y has a remainder of 5, what is 31 Jan 2010, 20:02
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Bunuel



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... :?
Show more

\(x=qy+5\), 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?
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If x and y are positive integers and x/y has a remainder of 5, what is 12 Feb 2010, 18:36
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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?

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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If x and y are positive integers and x/y has a remainder of 5, what is 02 Aug 2012, 17:45
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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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If x and y are positive integers and x/y has a remainder of 5, what is 02 Aug 2012, 19:17
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GmatSlayer112
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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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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If x and y are positive integers and x/y has a remainder of 5, what is 02 Aug 2012, 23:08
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GmatSlayer112
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.
Show more

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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If x and y are positive integers and x/y has a remainder of 5, what is Updated on: 08 Oct 2022, 22:19
Originally posted by KarishmaB on 02 Aug 2012, 23:32.
Last edited by KarishmaB on 08 Oct 2022, 22:19, edited 1 time in total.
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GmatSlayer112
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.
Show more


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.
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If x and y are positive integers and x/y has a remainder of 5, what is 04 Aug 2012, 18:33
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joyseychow
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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for this first time, i exactly though the same way as Bunuel...good teacher bunuel...thanks for this,,,,
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If x and y are positive integers and x/y has a remainder of 5, what is 12 Jan 2014, 05:11
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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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If x and y are positive integers and x/y has a remainder of 5, what is 12 Jan 2014, 06:06
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kiransaxena1988
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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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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If x and y are positive integers and x/y has a remainder of 5, what is 26 Sep 2016, 04:18
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joyseychow
If x and y are positive integers and x/y has a remainder of 5, what is the smallest possible value of xy?


Show Spoiler
This ques is from Man prep book. I don't understand why smallest value of x is 5 and not 11.
Show more
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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If x and y are positive integers and x/y has a remainder of 5, what is 26 Jan 2018, 08:10
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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
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If x and y are positive integers and x/y has a remainder of 5, what is 26 Jan 2018, 08:16
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MvArrow
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
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1 divided by 5 gives the remainder of 1, not 5.
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If x and y are positive integers and x/y has a remainder of 5, what is 26 Jan 2018, 08:54
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MvArrow
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.
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True! Today I have my mind in the clouds :dazed
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