Money Short

From the question, we can set up system of equations as shown:

$\dfrac{2}{3}(a-1) = b+1$

$\dfrac{3}{5}(b-2) = c+2$

$\dfrac{5}{7}(c-3) = d+3$

Solving all variables in terms of $a$ , we will get:

$b =\dfrac{2a-5}{3}$

$c =\dfrac{2a-21}{5}$

$d =\dfrac{2a-57}{7}$

Since $b, c, d$ are integers, we can set up equivalences of moduli $3, 5, 7$ :

$2a \equiv 5 \equiv 2 \pmod 3$

$2a \equiv 21 \equiv 1 \pmod 5$

$2a \equiv 57 \equiv 1 \pmod 7$

Suppose $x = 2a -1$ . Then $x \equiv 0 \pmod {35}$ or $x = 2a-1 = 35k$ for some odd integer $k$ . ( $k$ can't be even because $2a-1$ is odd.)

Then $x \equiv 35k \equiv 1 \pmod 3$ .

Hence, $k \equiv 2 \pmod 3$ .

Therefore, $k = 5 + 6n$ for some integer $n$ .

If $k = 5$ , then $2a-1 = 35k = 175$ or $a = 88$ .

If $k > 5$ , then the total money would exceed $200$ .

That means $a = 88 + 105m$ for some integer $m$ .

However, if $m > 0$ , the total money will exceed $200$ , which doesn't correlate with the question.

Therefore, $a = 88$ .

$b = \dfrac{2}{3}(88-1) -1 = 57$

$c = \dfrac{3}{5}(57-2) -2 = 31$

$d = \dfrac{5}{7}(31-3) -3 = 17$

As a result, $a+b+c+d = 88+57+31+17 =193$ , so they were $200-193= \boxed{7}$ dollars short.