Question

Let S = {4, 6, 9} and T = {9, 10, 11, …,1000}. If A = {a1 + a2 + … +ak :k∈N, a1, a2, a3, …, ak∈S}, then the sum of all the elements in the set T – A is equal to ____________.

Answer 1

Correct Answer: 11

The correct answer is 11
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