Vtyjkyuk

A group of soldiers was asked to fall in line making rows of three. It was found that there was one soldier extra. Then they were asked to stand in rows of five. It was found there were left two soldiers. They were asked to stand in rows of seven. Then there were three soldiers who could not be adjusted. At least how many soldiers were there in the group?

Now I know the answer to this is $52$. But how to arrive at it? How to explain it to my students?

Now I know when if the number of soldiers is $x$, then $x/3$ gives $1$ as remainder, $x/5$ gives $2$ as remainder and $x/7$ gives $3$ as remainder.

I can even explain that $x + 2$ is divisble by $2$, $x + 3$ is divisible by $5$ and $x + 4$ is divisible by $7$. But now what? How to arrive at the solution?

Start of by writing those equations in modular form like this:
$$begincases
xequiv 1pmod 3\
xequiv 2pmod5\
xequiv 3pmod7\
endcases$$

Then start of by writing the next congruence with the largest modulus namely $xequiv 3pmod 7$.

You can write this as $x=7*j + 3$ with $jinmathbb Z_0$.

Then you look for the next largest modulus $xequiv 2pmod 5$ which you can write as $j=5*k +2$ with $kinmathbb Z_0$

$qquad x= 7cdot(5cdot k+2)+3=35cdot k+17equiv 1pmod 3$

$beginaligniff 16-35cdot kequiv 0pmod 3\
kequiv 1pmod 3\endalign$

$beginalign
Longrightarrow x=35(3cdot l+1) +17\
x= 105cdot l+52
endalign$

with $linmathbb Z_0$.

So if the question wasn't "At least how many soldiers were there in the group" but "Give all the solutions to which the system of modular equations hold". It's $52+105cdot l$.