Let \(\mathrm{S}=\{4,6,9\}\) and \(\mathrm{T}=\{9,10,11, \ldots, 1000\}\). If \(\mathrm{A}=\left\{\mathrm{a}_{1}+\mathrm{a}_{2}+\ldots+\mathrm{a}_{\mathrm{k}}: \mathrm{k} \in \mathrm{N}, \mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{k}} \in \mathrm{S}\right\}\), then the sum of all the elements in the set \(\mathrm{T}-\mathrm{A}\) is equal to ______.

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Permutations and Combinations

Let

S = {4, 6, 9}

and

T = {9, 10, 11, \dots, 1000}

. If

A = {a_{1} + a_{2} + \dots + a_{k} : k \in N, a_{1}, a_{2}, a_{3}, \dots, a_{k} \in S}

, then the sum of all the elements in the set

T - A

is equal to ______.

Answer: 11

Set

A

consists of all positive integers that can be expressed as

4 x + 6 y + 9 z

where

x, y, z \in W

(non-negative integers).

By the Frobenius coin problem, any integer

\geq 6

except 11 can be represented as

4 x + 6 y

(since

g cd (4, 6) = 2

, the largest number not representable is

4 \cdot 6/2 - 4 - 6 = 2

, but we need odd numbers too). Including 9 allows representation of more numbers.

Checking: 9 = 9, 10 = 4+6, 12 = 4+4+4, etc. The only element in

T

that cannot be expressed as a sum of elements from

S

is 11 (since 11 = 4+4+3 is impossible with 4, 6, 9).

Therefore

T - A = {11}

, and the sum is 11.

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Integer answer