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The problem is:

$W$ is a positive integer when divided by $5$ gives remainder $1$ and when divided by $7$ gives remainder $5$. Find $W$.

Answer: Take the larger divisor , So Expression becomes $7k+5$. Now this number when divided by $5$ gives remainder $1$ so expression becomes $7k+4$ which is divisible by $5$. Now plugging in values for $k$ as $0,1,2,..$ we get $7(3)+4$ which is divisible by $5$ so $k=3$.

I know that $\frac{W}{5} => remainder~1$ so $5q+1=W$
$\frac{W}{7} => remainder~5$ so $7k+5=W$

Edit: Now I know that $5q+1 = 7k+5$

This implies $q=\frac{7k+4}{5}$

Now how did the text assume that $7k+4$ is divisible by $5$ and there would be no remainder?

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    $5k+1=7K+5 \Rightarrow 5k=7K+4$2012-07-16

2 Answers 2

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If $7k+5$ leaves a remainder of 1 when divided by 5, then $7k+4$ leaves a remainder of 0 when divided by 5, and so is divisible by 5.

To say that a number, say $7k+5$ "leaves a remainder of 1 when divided by 5" is exactly to say that it has the form $5q+1$ for some integer $q$. So we have $7k+5=5q+1$ for some integer $q$, and so $7k+4 = 5q$. Therefore $7k+4$ is a multiple of 5.

In general, of $x$ leaves a remainder of $r$ when divided by $n$, then $x-1$ leaves a remainder of $r-1$, unless $r=0$.

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    Yes thanks for the explanation.2012-07-16
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Hint $\rm\ \, 7k+5\, =\, 5\, q + 1\:\Rightarrow\: 7k+4\, =\, 5\,q,\ $ i.e. $\rm\:5\:|\:7k+4.$

Remark $\ $ You may find helpful my posts on Easy CRT.

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    Just added to my question..2012-07-16