"Let $S=\{4,6,9\}$ and $T=\{9,10,11, \ldots, 1000\}$. If $$A=\left\{a_{1}+a_{2}+\ldots+a_{k}: k \in \mathbf{N}, a_{1}, a_{2}, a_{3}, \ldots, a_{k}\right.$$ $\epsilon S\}$, then the sum of all the elements in the set $T-A$ is equal to __________."

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Here $S = \{ 4,6,9\}$

And $T = \{ 9,10,11,\,\,......,\,\,1000\}$.

We have to find all numbers in the form of $4x + 6y + 9z$, where $x,y,z \in \{ 0,1,2,\,......\}$.

If a and b are coprime number then the least number from which all the number more than or equal to it can be express as $ax + by$ where $x,y \in \{ 0,1,2,\,......\}$ is $(a - 1)\,.\,(b - 1)$.

Then for $6y + 9z = 3(2y + 3z)$

All the number from $(2 - 1)\,.\,(3 - 1) = 2$ and above can be express as $2x + 3z$ (say t).

Now $4x + 6y + 9z = 4x + 3(t + 2)$

$= 4x + 3t + 6$

again by same rule $4x + 3t$, all the number from $(4 - 1)\,(3 - 1) = 6$ and above can be express from $4x + 3t$.

Then $4x + 6y + 9z$ express all the numbers from 12 and above.

again 9 and 10 can be express in form $4x + 6y + 9z$.

Then set $A = \{ 9,10,12,13,\,....,\,1000\} .$

Then $T - A = \{ 11\}$

Only one element 11 is there.

Sum of elements of $T - A = 11$

"

Let $S=\{4,6,9\}$ and $T=\{9,10,11, \ldots, 1000\}$. If $$A=\left\{a_{1}+a_{2}+\ldots+a_{k}: k \in \mathbf{N}, a_{1}, a_{2}, a_{3}, \ldots, a_{k}\right.$$ $\epsilon S\}$, then the sum of all the elements in the set $T-A$ is equal to __________.

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Let $S=\{4,6,9\}$ and $T=\{9,10,11, \ldots, 1000\}$. If $$A=\left\{a_{1}+a_{2}+\ldots+a_{k}: k \in \mathbf{N}, a_{1}, a_{2}, a_{3}, \ldots, a_{k}\right.$$ $\epsilon S\}$, then the sum of all the elements in the set $T-A$ is equal to __________.

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