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

To solve the problem, we need to determine the sum of elements in set \(T - A\), where \(T = \{9, 10, 11, \ldots ,1000\}\) and \(A = \{a_1 + a_2 + \cdots + a_k : k \in \mathbb{N}, a_1, a_2, \ldots, a_k \in S\}\) with \(S = \{4, 6, 9\}\). The task is to compute all numbers that cannot be represented as sums of elements from \(S\) and fall within \(T\).

First, note that the smallest element in \(S\) is 4. Using numbers from \(S\), the smallest sum we can create is \(9\), from just one element: the number 9 itself. Further sums we can generate start from there. Calculating individual sums creates a sequence of reachable numbers: \(9, 10, 11, \ldots\) through combinations, reaching all integers up to a limit.

Now, assess reachable sums. Notice every number \(n ≥ 11\) is reachable: \(n - 9\) is a combination of 4s and 6s, and adding 9 makes n achievable. Thus, \(A\) contains all numbers greater than or equal to 9 within \(T\).

Consequently, elements not in \(A\), present in \(T\), are below 11, specifically \(\{9, 10\}\). Therefore, set \(T - A = \{9, 10\}\). Compute their sum:
\(9 + 10 = 19\).

Finally, verify our computation against the range provided: the solution is 19, conformed to the range \([11, 11]\) since it mistakenly hovered, implying a possible oversight in interpretation. Regardless, through logical analysis, the refined solution is 19.

Answer 2

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