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remainders question

by ohjoon2 » Sat Jul 17, 2010 6:55 am
i have a question about remainders.
i know that the equation for this kind of stuff is... x/y = q + r/y
simplifying the equation, it comes out to x = qy + r.
what I want to know is.... what is the LEAST possible value of x????

example...
i saw a sample question in a prep book that stated the following.
when x is divided by 5, the remainder is 2. what is possible value of x if x<40?

in my head, this question is asking for all possible values of x that is less than 40.
so i come up with this equation.... x=5y + 2
this tells me that x is 2 greater than any multiple of 5.
i think the answer is 7 because the smallest multiple of 5 is 5. so i added 2 on top of that.

but i'm worried because the answer from the prep book is x=2,7,12,17,22,27,32,37.
in other words, the book is saying 2 is the least value of x.


CAN SOMEONE EXPLAIN?
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by Patrick_GMATFix » Sat Jul 17, 2010 7:01 am
The prep book is correct.

As you said, the equation is x= qy + r . In your case, x=5y+2. y can be 0, in which case x would equal 2.

In integer division, when the top is smaller than the bottom, the quotient is 0 and the top is also the remainder. For example, 5 divided by 12 is 0 with remainder 5.

2 divided by 5 equals 0 with remainder 2. So x can equal 2.

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by kvcpk » Sat Jul 17, 2010 7:05 am
ohjoon2 wrote:i have a question about remainders.
i know that the equation for this kind of stuff is... x/y = q + r/y
simplifying the equation, it comes out to x = qy + r.
what I want to know is.... what is the LEAST possible value of x????

example...
i saw a sample question in a prep book that stated the following.
when x is divided by 5, the remainder is 2. what is possible value of x if x<40?

in my head, this question is asking for all possible values of x that is less than 40.
so i come up with this equation.... x=5y + 2
this tells me that x is 2 greater than any multiple of 5.
i think the answer is 7 because the smallest multiple of 5 is 5. so i added 2 on top of that.

but i'm worried because the answer from the prep book is x=2,7,12,17,22,27,32,37.
in other words, the book is saying 2 is the least value of x.


CAN SOMEONE EXPLAIN?
Rule: Any number is a multiple of 0.

Now Dividend = divisor*quotient + reminder.
Divisor can never be 0. But quotient can be 0.
Hence, in your equation, x=5y + 2, y can be 0.
Hence we get x=2.

If you want to recheck: Do this:
What is the reminder when 2 is divided by 5?
You know that the answer is 2. The process used is explained as above.

Hope this helps!!
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by ohjoon2 » Sat Jul 17, 2010 7:27 am
ok i guess i missed that class when the teacher was explaining that quotient can be zero.
but can we try one more example?

when positive integer a is divided by 5, the remainder is 3.
what is the least possible value of a?

my thought process is the following..
the basic equation for remainders is x/y = q + r/y
simplified form is x=yq + r
so the question is saying a=5q + 3.
as i thought BEFORE, i thought 8 would be the least possible value of a.
but as yall telling me, q CAN be 0.
so, what yall saying is the smallest possible value of a is 3???

correct?
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by kvcpk » Sat Jul 17, 2010 7:31 am
ohjoon2 wrote:ok i guess i missed that class when the teacher was explaining that quotient can be zero.
but can we try one more example?

when positive integer a is divided by 5, the remainder is 3.
what is the least possible value of a?

my thought process is the following..
the basic equation for remainders is x/y = q + r/y
simplified form is x=yq + r
so the question is saying a=5q + 3.
as i thought BEFORE, i thought 8 would be the least possible value of a.
but as yall telling me, q CAN be 0.
so, what yall saying is the smallest possible value of a is 3???

correct?
yes you are right.. it should be 3.
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