Pay it up!

We could consider all the ways to evaluate $442$ using $\{1,7,23,59\}$ like

$\displaystyle a+b\cdot 7+c \cdot 23 + d \cdot 59 =442$

where

$\displaystyle 0\leq a\leq \left \lfloor{\frac{442}{1}} \right \rfloor=442$

$\displaystyle 0\leq b\leq \left \lfloor{\frac{442}{7}} \right \rfloor=63$

$\displaystyle 0\leq c\leq \left \lfloor{\frac{442}{23}} \right \rfloor=19$

$\displaystyle 0\leq d\leq \left \lfloor{\frac{442}{59}} \right \rfloor=7$ .

The right-side bound is the biggest coefficient such that I can multiplay $a,b,c$ or $d$ before finding a number greater than $442$ . Let's use a simple MATLAB code to evaluate all the possible occurrences:

$\text{RationalQ}=\text{Simplify}[\text{\$\#\$1}\in \mathbb{Q}]\&;$

$\text{change}( \\ \text{amount\$\_\$}\text{/;}\text{RationalQ}(\text{amount})\land \text{amount}\geq 0, \\ \text{currency\$\_\_\$}\text{/;} \\ \text{Head}[\text{currency}]=\text{List}\land \\ \text{AllTrue}[\text{currency},\text{RationalQ}]\land \\ \text{AllTrue}[\text{currency},\text{Positive}])\text{:=} \\ \text{Block}[\{c,a,s\}, \\ c(0,\_)\text{:=}1; \\ c(\text{a\$\_\$},\_)\text{:=}0\text{/;}a<0; \\ c(\text{a\$\_\$},\{\})\text{:=}0\text{/;}a\neq 0; \\ c(\text{a\$\_\$},\text{s\$\_\$})\text{:=}c(a,s)=c(a-\text{First}[s],s)+c(a,\text{Rest}[s]); \\ c(\text{amount},\text{Sort}[\text{Union}[\text{currency}],\text{Greater}])];$

$\text{change}(442,\{59,23,7,1\}) \Longrightarrow 2018$

$\sum_{a=0}^7 \sum_{b=0}^{\lfloor\frac{1}{23}(442 - 59a)\rfloor}\left(1 + \big\lfloor \tfrac17(442 - 59a - 23b)\big\rfloor\right) \; = \; \boxed{2018}$ counts the number of ways of paying with $a \times 59\mathbb{B}$ , $b \times 23\mathbb{B}$ , $c \times 7\mathbb{B}$ , with the rest in singles.