Let S = {4, 6, 9} and T = {9, 10, 11,…., 1000}.

If A = $\left\{a_1+a_2+...+a_k:k\in N,a_1,a_2,a_3,.....,a_k\in S\right\},$ then the sum of all the elements in the set T – A is equal to…………….

<p>T = {9, 10, 11, 12, …., 1000}</p><p>S = {4, 6, 9}</p><p>A =&nbsp;<math><mrow><mrow><mo> {</mo><mrow><msub><mrow><mi>a</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>+</mo><msub><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>+</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>+</mo><msub><mrow><mi>a</mi></mrow><mrow><mi>k</mi></mrow></msub><mo>:</mo><mi>k</mi><mo>∈</mo><mi>N</mi><mo>, </mo><msub><mrow><mi>a</mi></mrow><mrow><mi>i</mi></mrow></msub><mo>∈</mo><mi>S</mi></mrow><mo>}</mo></mrow></mrow></math></p><p>Let 4 appear x no. of times</p><p>6 appear y no. of times</p><p>9 appear z no. of times</p><p>10 then set A has n element 4x + 6y + 9z</p><p>Concept : Let a and b are co-prime numbers, then members from (a - ) (b – 1) and more can be expressed in the form ax + by where x, y&nbsp;<math><mrow><mo>∈</mo></mrow></math>&nbsp; {0, 1, 2, …}</p><p>So, all the number of the form</p><p>2y + 3z are (2 1). (3 – 1), ….</p><p>i.e., 6, 7, 8, 9, 10, 11, ….</p><p>form : 2 + t, t = 0, 1, 2, 3, ….</p><p>So, 6y + 9z = 3 (2y + 3z) = 3 (2 + t) = 6 + 3t, t = 0, 1, 2, 3, …</p><p>sum of element of T – A is 11</p>

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