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Divisibility by 7

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chitrangada.maitra
 Post subject: Divisibility by 7
 Post Posted: Tue Aug 03, 2010 6:20 pm 
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Course Students


Posts: 75
This problem has been troubling me for a while.
Strategy Guide: Number Properties, page 134, #20

x has a remainder of 5 when divided by 9. y has a remainder of 7 when divided by 9. what is the remainder when x-y is divided by 9

My Approach:
X can be 5, 14, 23, 32, 41... etc
Y can be 7, 16, 25, 34, 43... etc

If X>y, the remainder of X-Y upon division by 9 is consistently 7.
However, if, X<Y, X-Y can be -2, -11, -20.... etc
These numbers upon division by 9 does NOT give one consistent answer.

The strategy guide suggests adding the divisor to a negative remainder -- that is the part that confuses me.

Can someone please suggest a simpler way to solve this.

Thanks,
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gokul_nair1984
 Post subject: Re: Divisibility by 7
 Post Posted: Wed Aug 11, 2010 1:21 am 
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Students


Posts: 170
You are making a fundamental error in divison. Don't fret it is a very common one.

The first statement which says that if X>Y, the remainder when X-Y is divided by 9 will always be 7 is perfect.

Let us consider the case when X<Y. Even then the remainder will consistently be 7. Here's how this works.

Most conspicuously, always remember that Divisor*Quotient should always be lesser than Dividend for subtraction to take place.
The mistake you are making here is that you are considering 0 to the quotient when -2 is divided by 9, which in turn needs you to subtract 0 from -2(this is not possible as 0>-2). Thus the quotient has to be -1, which inturn needs you to subtract -9 from -2 =>-2-(-9)=7.
Here, you can subtract since -9<-2.

I hope this concept is clear..
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