Solution:
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∈{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∈{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
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