1. Interpret the set $A$

We have $S=\{4,6,9\}$ and $A=\{a_1+a_2+\cdots+a_k: k\in \mathbb N,\ a_i\in S\}.$

So $A$ is the set of all positive integers that can be written as a sum of one or more elements from $\{4,6,9\}$.

We need the sum of all elements in $T-A,$ where $T=\{9,10,11,\dots,1000\}.$

Thus we must find all integers from $9$ to $1000$ that cannot be expressed as a sum of $4,6,9$.

2. Find which numbers are representable

Let us test small integers starting from $9$.

• $9=9$ so $9\in A$
• $10=4+6$ so $10\in A$
• $11$ cannot be written as a sum of $4,6,9$
• $12=6+6=4+4+4$ so $12\in A$
• $13=9+4$ so $13\in A$
• $14=4+4+6$ so $14\in A$
• $15=9+6$ so $15\in A$
• $16=4+6+6=4+4+4+4$ so $16\in A$

Now observe an important fact: if some number $n\in A$, then $n+4\in A$ as well, because we can just add another $4$.

Since we have shown that $12,13,14,15\in A,$ all numbers greater than or equal to $12$ are in $A$. Indeed:

• from $12$ we get $16,20,24,\dots$
• from $13$ we get $17,21,25,\dots$
• from $14$ we get $18,22,26,\dots$
• from $15$ we get $19,23,27,\dots$

These cover every integer $\ge 12$.

So among integers in $T=\{9,10,11,\dots,1000\}$, the only one not in $A$ is $11.$

Hence, $T-A=\{11\}.$

3. Sum of all elements of $T-A$

The sum is simply $11.$

4. Compare with stored correct answer

Stored correct answer = $11$.

This matches our derived answer.