Q:

Calculation of integers $b,c,d,e,f,g$ such that $\frac{5}{7} = \frac{b}{2!}+\frac{c}{3!}+\frac{d}{4!}+\frac{e}{5!}+\frac{f}{6!}+\frac{g}{7!}$

There are unique integers $b,c,d,e,f,g$ such that $\displaystyle \frac{5}{7} = \frac{b}{2!}+\frac{c}{3!}+\frac{d}{4!}+\frac{e}{5!}+\frac{f}{6!}+\frac{g}{7!}$
Where $0\leq b,c,d,e,f,g <i$ for $i=2,3,4,5,6,7$. Then the value of $b+c+d+e+f+g = $
$\bf{My\; Try}::$ $\displaystyle \frac{5}{7} = \frac{2520\cdot b+840\cdot c+210\cdot d+42\cdot e+7\cdot f+g}{7\times 720}$
$\displaystyle 2520\cdot b+840\cdot c+210\cdot d+42\cdot e+7\cdot f+g = 720\times 5 = 3600$
Now I did not understand how can I solve after that.
Help Required
Thanks

A:

Look at this equality that you've got: $$ 2520 \cdot b + 840 \cdot c + 210 \cdot d + 42 \cdot e + 7 \cdot f + g = 3600.$$
Note that if you consider everything modulo $7$, then most of the summands disappear, because $2520,840,210,42$ and $7$ are all multiples of $7$. So, taking remainders modulo $7$, we get $g \equiv 2 \pmod 7$. Since $0 \leq g < 7$, it follows that $g = 2$. Now substitute $2$ for $g$ in your equality, subtract $2$ from both sides and divide everything by $7$. You get
$$
360 \cdot b + 120 \cdot c + 30 \cdot d + 6 \cdot e + f = 514.
$$
Now consider both sides modulo $6$, and go on in a similar fashion. You will eventually find the values for all variables.

