I am trying to solve this problem
W is a positive integer when divided by 5 gives remainder 1 and when divided by 7 gives remainder 5. Find W.
I am referring back to an earlier post I made. Now I am attempting to solve it that way.
We know that
$$w\equiv1(mod~5)$$
$$w\equiv5(mod~7)$$
$w=7r+5$
$w=5r+5+2r$
Since $5r+5$ is divisible by 5
$w=5(r+1)+2r$ this shows the remainder is $2r$
Now $2r$ divided by 5 gives a remainder 1 , thus giving the equation
$2r = 5k + 1$ or $r=\frac{5k+1}{2}$
Putting r back in $w=7r+5$ we get
$2w = 35k + 12$
So I guess $w= \frac{35+12}{2} = 23.5$
This is wrong and the answer is suppose to be 26. Any suggestions what I might be doing wrong ? or anything that I might be missing ?
Edit:
The problem was in calculation
$w= \frac{35+17}{2} = 26$
Best Answer
There is an error: $\rm\:w=7r\!+\!5\,\Rightarrow\,2w = 7(2r)\!+\!\color{#C00}{10} = 7(5k\!+\!1)\!+\!\color{#C00}{10} = 35k\!+\!\color{#C00}{17},\:$ not $\rm\:35k\!+\!\color{#0A0}{12}.$ Since $\rm\:35k\!+\!17 = 2w\:$ is even, $\rm\:k\:$ is odd, $\rm\:k = 2j\!+\!1,\,$ so $\rm\:w = (35(2j\!+\!1)\!+\!17)/2 = 35j+26.$
Remark $\ $ It is easier to do the division by $2$ before the substitution. Namely, we have $\rm\:2r = 5k\!+\!1\:$ so $\rm\:k\:$ is odd, $\rm\:k = 2j\!+\!1,\:$ thus $\rm\:r = (5k\!+\!1)/2 = (5(2j\!+\!1)\!+\!1)/2 = 5j\!+\!3.\:$ Therefore $\rm\:w = 7r\!+\!5 = 7(5j\!+\!3)\!+\!5 = 35j\!+\!26.$ Notice how the numbers remain smaller this way.
I emphasize again, it's much more intuitive if you learn about modular arithmetic (congruences). For many examples see my posts on Easy CRT (easy version of the Chinese Remainder Theorem)