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another remainder problem

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another remainder problem

by bfman » Thu Jul 30, 2009 6:59 pm
Q: If x and y are positive integers and x/y has a remainder of 5, what is the smallest possible value of xy?

I got 66, but book's answer is 30. They made the smallest possible value of x to be 5. I don't understand how they did this.

I agree that y has to be 6, but in order to get a remainder of 5, doesn't x have to be 11? and hence 6*11 = 66?

Thanks.
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by scoobydooby » Thu Jul 30, 2009 7:27 pm
x=ky+5 where k is a constant,

minimum value of x is when k=0 (given x is a positive integer)
=>minimum value of x=5
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by bfman » Thu Jul 30, 2009 8:05 pm
I am sorry but I have no clue what you did there.... lol
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by scoobydooby » Thu Jul 30, 2009 8:50 pm
x=ky+r is an equation where x=dividend, y=divisor, r=remainder, k=integral quotient

when we divide 8 by 2 we get 4, remainder:0
=>8=2*4+0

when we divide 8 by 3 we get remainder: 2
=>8=3*2+2

in the question above, it is given that the remainder is 5,
so x=ky+5
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by tohellandback » Thu Jul 30, 2009 8:52 pm
X=nY +5 from the question
n>=0
and y>5 to get a remainder 5
we need to find the minimum value of XY. minimum value of Y is 6
to find the minimum value of X, start with n=0, i.e. X=5
so answer is 6*5=30
The powers of two are bloody impolite!!
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