<s>[INST] Elementary number theory, 'concert-ticket-arithmetic' Four friends, call them A,B,C and D are planning to go to a concert, but they realize that they 
        are a few dollars short to buy tickets.(50 $ per ticket). 
   We know that each of them has an integer amount of dollars.

   If B  borrowed 1$ from A,then  B   would have  2/3 of A’s balance

   If C  borrowed 2$ from B,then  C   would have  3/5 of B’s balance

    If D  borrowed 3$ from C, then  D   would have  5/7  of C’s balance

At least how much more money do they need(in $) in order to afford 4 tickets?
 [/INST] Let $a$ be the money of $A$. 
$b$ be the money of $B$. $c$ the money of $C$ and $d$ the money of $D$. I have the following system: 
$$\left\{\begin{matrix}
b+1=\frac{2}{3}(a-1)
\\c+2=\frac{3}{5}(b-2)
\\d+3=\frac{5}{7}(c-3)
\end{matrix}\right.$$
Solving for $a,b,c,d$, I obtain: 
$$\left\{\begin{matrix}
b=\frac{2}{3}(a-1)-1
\\c=\frac{1}{5}(2a-1)-4
\\d=\frac{1}{7}(2a-1)-8
\end{matrix}\right.$$
From this I can say: $3|a-1
\land 7|2a-1 \land 5|2a-1$ or in other words: $3|a-1\land 35|2a-1$. Also: 
$$\left\{\begin{matrix}
35\alpha=2a-1
\\3\beta=a-1
\end{matrix}\right.$$
With $\alpha,\beta \in Z$. Solving for $\alpha$ and $\beta$, I obtain: 
$$\left\{\begin{matrix}
\alpha=6n+5
\\\beta=35n+29
\end{matrix}\right.$$
With $n \in N$, So, setting $n=0$, I obtain the solutions:
$$\left\{\begin{matrix}
a=88
\\b=57
\\c=31
\\d=17
\end{matrix}\right.$$
Now let $R$ the money they need: $R=200-a-b-c-d=200-88-57-31-17=7$.
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