Animated Solution for Mathematics - Sets and Relations: Let $S=\{4,6,9\}$ and $T=\{9,10,11,\dots ,1000\}$. If $A=\{a_1+a_2+\cdots +a_k:k\in \mathbb{N},a_1,a_2,\dots ,a_k\in S\}$, then the sum of all the elements in the set $T-A$ is equal to ______.

We are investigating the set $S=\{4,6,9\}$ and the target range $T=\{9,10,11,\dots ,1000\}$. We aim to identify which integers in $T$ cannot be expressed as a linear combination of the form:

where $x,y,z\in \{0,1,2,\dots \}$ and at least one variable is positive. This is a variation of the Frobenius Coin Problem.

Let us evaluate the reachability of the first few integers in $T$:

For $n=9$: We have $9(1)$, so $9\in A$.

For $n=10$: We have $4(1)+6(1)=10$, so $10\in A$.

For $n=11$: If we use a $9$, we require $2$ more, which cannot be formed by $4$ and $6$. If we exclude $9$, we must solve $4x+6y=11$. Since $4x+6y$ is always even for non-negative integers $x$ and $y$, it can never equal the odd number $11$. Thus, $11 otin A$.

Now, let us examine the integers starting from $12$:

We have identified a streak of four consecutive integers $\{12,13,14,15\}$ that all belong to $A$.

Because the smallest generator in our set $S$ is $4$, once we establish a streak of $4$ consecutive integers, we can generate any subsequent integer $n>15$ by adding $4$ to one of the numbers in the streak. This is the Principle of Continuity.

By establishing this streak, we have proven that every integer $n\ge 12$ is contained in $A$.

We previously verified that $9\in A$ and $10\in A$. The only integer in the range $T$ that cannot be formed is $11$.

Therefore, the set of unreachable numbers in $T$ is $\{11\}$. The sum of all elements in $T-A$ is $11$.