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

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Intern
Joined: 02 Sep 2016
Posts: 21
Re: If x and y are positive integers and x/y has a remainder of  [#permalink]

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26 Sep 2016, 03:18
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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Joined: 06 Sep 2016
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Location: Italy
Schools: EDHEC (A\$)
GMAT 1: 650 Q43 V37
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Re: If x and y are positive integers and x/y has a remainder of  [#permalink]

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26 Jan 2018, 07:10
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: 52296
Re: If x and y are positive integers and x/y has a remainder of  [#permalink]

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26 Jan 2018, 07:16
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: 132
Location: Italy
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  [#permalink]

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26 Jan 2018, 07:54
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 &nbs [#permalink] 26 Jan 2018, 07:54

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