JEE Main 2022 — Sets And Relations Question with 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 $$x,y,z\in \{0,1,2,......\}$$.

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 $$T-A=11$$