jmc

Suppose we require $a$ $7$s, $b$ $77$s, and $c$ $777$s to sum up to $7000$ ($a,b,c\ge 0$). Then $7a+77b+777c=7000$, or dividing by $7$, $a+11b+111c=1000$. Then the question is asking for the number of values of $n=a+2b+3c$. Manipulating our equation, we have $a+2b+3c=n=1000-9(b+12c)⟹0\le 9(b+12c)<1000$. Thus the number of potential values of $n$ is the number of multiples of $9$ from $0$ to $1000$, or $112$. However, we forgot to consider the condition that $a\ge 0$. For a solution set $(b,c):n=1000-9(b+12c)$, it is possible that $a=n-2b-3c<0$ (for example, suppose we counted the solution set $(b,c)=(1,9)⟹n=19$, but substituting into our original equation we find that $a=-10$, so it is invalid). In particular, this invalidates the values of $n$ for which their only expressions in terms of $(b,c)$ fall into the inequality $9b+108c<1000<11b+111c$. For $1000-n=9k\le 9(7\cdot 12+11)=855$, we can express $k$ in terms of $(b,c):n\equiv b(\mathrm{mod}12),0\le b\le 11$ and $c=\frac{n-b}{12}\le 7$ (in other words, we take the greatest possible value of $c$, and then "fill in" the remainder by incrementing $b$). Then $11b+111c\le 855+2b+3c\le 855+2(11)+3(7)=898<1000$, so these values work. Similarily, for $855\le 9k\le 9(8\cdot 12+10)=954$, we can let $(b,c)=(k-8\cdot 12,8)$, and the inequality $11b+111c\le 954+2b+3c\le 954+2(10)+3(8)=998<1000$. However, for $9k\ge 963⟹n\le 37$, we can no longer apply this approach. So we now have to examine the numbers on an individual basis. For $9k=972$, $(b,c)=(0,9)$ works. For $9k=963,981,990,999⟹n=37,19,10,1$, we find (using that respectively, $b=11,9,10,11+12p$ for integers $p$) that their is no way to satisfy the inequality $11b+111c<1000$. Thus, the answer is $112-4=\boxed{108}$.