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If x and y are positive integers and x/y has a remainder of

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If x and y are positive integers and x/y has a remainder of [#permalink] New post 20 Jan 2010, 19:54
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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?


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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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Re: Remainder of 5 [#permalink] New post 20 Jan 2010, 21:32
As the remainder is 5 the numerator has to be a multiple of 5. Since only multiples of 5 have exactly 5 units between each one of them(5,10,15,20..). Now to tackle the denominator...look for the smallest possible positive integer. If we take 1, we quickly realize that it completely divides any number (remainder is 0).
So we go to the next integer which is 2 and we can conclude the multiple of 5 that should be in the numerator is 5x2=10.
Thus we can see that the smallest possible value is 10 x 2=20.
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Re: Remainder of 5 [#permalink] New post 21 Jan 2010, 00:49
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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: Remainder of 5 [#permalink] New post 30 Jan 2010, 14:45
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???? :shock: :roll: What is the best approach to master such things??? Please guide!

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Re: Remainder of 5 [#permalink] New post 30 Jan 2010, 14:59
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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???? :shock: :roll: 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: Remainder of 5 [#permalink] New post 31 Jan 2010, 07:26
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: Remainder of 5 [#permalink] New post 31 Jan 2010, 07:43
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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: Remainder of 5 [#permalink] New post 31 Jan 2010, 19:02
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?
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Re: Remainder of 5 [#permalink] New post 01 Feb 2010, 04:17
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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:

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.

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: Remainder of 5 [#permalink] New post 12 Feb 2010, 17:36
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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MGMAT Remainder Question [#permalink] New post 18 Aug 2010, 16:31
i need a better explanation for their solution please.

Problem:
If x and y are positive integers and x/y has a remainder of 5, what is the smallest possible value of xy?

solution:
30: The remainder must always be smaller than the divisor. In this problem, 5 must be smaller than y. Additionally, y must be an integer, so y must be at least 6. If y is 6, then the smallest possible value of x is 5. (Other values of x that leave a remainder of 5 when divided by 6 would be 11, 17, 23, etc.) If y is chosen to be larger than 6, then the smallest possible value of x is still 5. Thus, we will get the smallest possible value of the product xy by choosing the smallest x together with the smallest y. The smallest possible value of xy is 5X6=30.

My problem, is that I do not know why the smallest possible value of x is 5. My logic is that since the remainder is 5, y is greater than 5 and is in turn 6. So for a number to yield a remainder of 5 after being divided by 6, 5+6=11. X must be 11 and xy=66. Dont see how you can have 5/6 has a reminader of 5.

thanks
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Re: MGMAT Remainder Question [#permalink] New post 18 Aug 2010, 18:04
chsebik wrote:
i need a better explanation for their solution please.

Problem:
If x and y are positive integers and x/y has a remainder of 5, what is the smallest possible value of xy?

solution:
30: The remainder must always be smaller than the divisor. In this problem, 5 must be smaller than y. Additionally, y must be an integer, so y must be at least 6. If y is 6, then the smallest possible value of x is 5. (Other values of x that leave a remainder of 5 when divided by 6 would be 11, 17, 23, etc.) If y is chosen to be larger than 6, then the smallest possible value of x is still 5. Thus, we will get the smallest possible value of the product xy by choosing the smallest x together with the smallest y. The smallest possible value of xy is 5X6=30.

My problem, is that I do not know why the smallest possible value of x is 5. My logic is that since the remainder is 5, y is greater than 5 and is in turn 6. So for a number to yield a remainder of 5 after being divided by 6, 5+6=11. X must be 11 and xy=66. Dont see how you can have 5/6 has a reminader of 5.

thanks


Quote:
So for a number to yield a remainder of 5 after being divided by 6


Think about this. If x is 5 and y is 6, then when 5 is divided by 6, the quotient is 0 and the remainder is 5.

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Remainder question [#permalink] New post 02 Aug 2012, 16:45
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: Remainder question [#permalink] New post 02 Aug 2012, 18:17
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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: Remainder question [#permalink] New post 02 Aug 2012, 22:08
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: Remainder question [#permalink] New post 02 Aug 2012, 22:32
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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: Remainder question [#permalink] New post 04 Aug 2012, 09:32
VeritasPrepKarishma wrote:
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/


Thanks very helpful, especially the blog post.
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Re: Remainder of 5 [#permalink] New post 04 Aug 2012, 17:33
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: Remainder of 5 [#permalink] New post 12 Jan 2014, 04:11
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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Kudos [?]: 22263 [0], given: 2602

Re: Remainder of 5 [#permalink] New post 12 Jan 2014, 05:06
Expert's post
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: Remainder of 5   [#permalink] 12 Jan 2014, 05:06
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