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Question about remainders

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Question about remainders  [#permalink]

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New post 03 Sep 2017, 14:51
I am going through Bunuel's compilation tips of remainders. I am a bit confused about one point and can't seem to get it. Any help would be appreciated.

3) If the value of ‘r’ is greater than the value of the factor, then we have to take the remainder of ‘r’
divided by the factor to get the remainder.
Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 3 (which is a factor of
21) will be remainder of 5/3, which is 2.

I understand that 26/21 will give us a remainder of 5 and 26/7 will also give us a remainder of 5. But wouldn't 26/3 also give us a remainder of 5 even thou r(5)>value of the factor (3)? I am confused about this tip and do not fully understand it. If someone could explain it in detail and provide examples that would greatly be appreciated.
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Re: Question about remainders  [#permalink]

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New post 03 Sep 2017, 17:55
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Hi Khalid,

Couple of things on Remainders.
1. If you are diving a number by "n" then you can get remainders only from 0 to "n-1"
2. Diving a number by a number say 4 is same as splitting the number into two parts as dividing the sub-parts by 4
Ex: 21/4 remainder is 1
21/4 = (15 + 6)/4 = (15/4) + (6/4)
= Remainder of 3 + remainder of 2 = total remainder of 5.
But we were diving the numbers by 4 so remainder cannot be greater than 4. So final remainder is 5/4 which is 1

Try taking more examples like this and try and this should be clear. Also, I have written a post on remainder sometime back which might help!
https://gmatclub.com/forum/how-to-solve ... 42591.html

khalid228 wrote:
I am going through Bunuel's compilation tips of remainders. I am a bit confused about one point and can't seem to get it. Any help would be appreciated.

3) If the value of ‘r’ is greater than the value of the factor, then we have to take the remainder of ‘r’
divided by the factor to get the remainder.
Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 3 (which is a factor of
21) will be remainder of 5/3, which is 2.

I understand that 26/21 will give us a remainder of 5 and 26/7 will also give us a remainder of 5. But wouldn't 26/3 also give us a remainder of 5 even thou r(5)>value of the factor (3)? I am confused about this tip and do not fully understand it. If someone could explain it in detail and provide examples that would greatly be appreciated.

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Re: Question about remainders  [#permalink]

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New post 04 Sep 2017, 08:36
khalid228 wrote:
I am going through Bunuel's compilation tips of remainders. I am a bit confused about one point and can't seem to get it. Any help would be appreciated.

3) If the value of ‘r’ is greater than the value of the factor, then we have to take the remainder of ‘r’
divided by the factor to get the remainder.
Eg. If remainder of a number when divided by 21 is 5, then the remainder of that same number when divided by 3 (which is a factor of
21) will be remainder of 5/3, which is 2.

I understand that 26/21 will give us a remainder of 5 and 26/7 will also give us a remainder of 5. But wouldn't 26/3 also give us a remainder of 5 even thou r(5)>value of the factor (3)? I am confused about this tip and do not fully understand it. If someone could explain it in detail and provide examples that would greatly be appreciated.


Pay attention to the highlighted part:

If \(x\) and \(y\) are positive integers, there exist unique integers \(q\) and \(r\), called the quotient and remainder, respectively, such that \(y =divisor*quotient+remainder= xq + r\) and \(0\leq{r}<x\).

For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since \(15 = 6*2 + 3\).

Notice that \(0\leq{r}<x\) means that remainder is a non-negative integer and always less than divisor.

So, when you divide by 3, the remainder could only be 0, 1, or 2. 26 divided by 3 gives the remainder of 2: 26 = 8*3 + 2.

For more on this please check the links below:

6. Remainders



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Re: Question about remainders   [#permalink] 04 Sep 2017, 08:36
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