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

Solution

Maths sir · 4 steps

Find a run of four consecutive sums, because adding 4 again and again then reaches every larger number.

1. What A holds. A is every number you can build by adding 4s, 6s and 9s, with repeats allowed. T - A is the numbers from 9 to 1000 that cannot be built this way.

   Number	Built from 4, 6, 9?
   9	9
   10	4 + 6
   11	no
   12	6 + 6
   13	4 + 9
   14	4 + 4 + 6
   15	6 + 9

2. Why 11 fails. Sums of 4s and 6s are always even, so an odd number like 11 needs at least one 9. Then 11 - 9 = 2 is left over, and 2 cannot be made from 4, 6 or 9.
3. Everything after 11. 12, 13, 14 and 15 are all in A. Adding 4 to these gives 16 to 19, adding 4 again gives 20 to 23, and so on. So every number from 12 up to 1000 is in A.
4. So. T - A = {11}, and the sum of its elements is 11.

The idea to keep. If you can reach m consecutive numbers, where m is the smallest number you are allowed to add, then every number beyond them is reachable too. Only check the small numbers before that run.

Answer from the NTA final answer key. Solution by PrepWiser.