Is it distinct 2? (Are the Solutions to this Problem Distinct?)

Let the 4 positive integers be $a$ , $2a$ , $3a$ and $b$ , then we have:

$\begin{aligned} \frac 1a + \frac 1{2a} + \frac 1{3a} + \frac 1b & = \frac {19}{20} \\ \frac {11}{6a} + \frac 1b & = \frac {19}{20} \\ \frac {11b+6a}{6ab} & = \frac {19}{20} \\ 110b + 60a & = 57ab \\ 57ab - 60a - 110b & = 0 \\ \left(3a - \frac {110}{19}\right)(19b-20) & = \frac {2200}{19} \\ (57a-110)(19b-20) & = 2200 \end{aligned}$

For the LHS to be even as the RHS, both $a$ and $b$ must be even. When $a=2$ , we have $b=30$ . Therefore, the four positive integers are 2, 4, 6 and 30 and their sum is $\boxed{42}$ .

Let the numbers be $x \ , \ 2x \ , \ 3x \ , \ y$

So, we are required to solve for $6x + y$ . Then,

$\displaystyle \begin{aligned} \dfrac{1}{x} + \dfrac{1}{(2x)} + \dfrac{1}{(3x)} + \dfrac{1}{y} & = \cfrac{19}{20} \\ \implies \dfrac{11}{(6x)} + \dfrac{1}{y} & = \dfrac{19}{20} \\ \dfrac{11}{(6x)} & = \dfrac{(19y - 20)}{(20y)} \\ \implies x & = \dfrac{(10\cdot11\cdot y)}{(3\cdot(19y-20))} \end{aligned}$

As $10\cdot11\cdot y \mod 3 \equiv 0 \implies y \mod 3 \equiv 0$ .

Let

$\displaystyle \begin{aligned} y &= 3k \\ x &= \dfrac{(10*11*k)}{(57k - 20)} \\ \implies x &= \dfrac{(10k)}{\dfrac{57k}{11} - \dfrac{20}{11}} \end{aligned}$

By using modulo,

$\displaystyle \begin{aligned} (57k \mod 11) - (20 \mod 11) & \equiv 0 \\ (2k \mod 11) - (20 \mod 11) & \equiv 0 \\ \implies k & = 10 \end{aligned}$

So, $y = 30 \implies x = 2$

We need to find $6x+y \implies (6\cdot2) + 30 = \boxed{42}$