Arithmetical set
{ 1 , 4 , 7 , 1 0 , 1 3 , 1 6 , 1 9 }
How many different integers can be expressed as the sum of three distinct members of the set above?
The answer is 13.
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1 solution
i don't understand why the answer is not 7C3. Maybe i misunderstand the question is, would you explain it?
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7 C 3 = 3 5 is the total number of combinations. But 2 2 of these combinations are repeated. For instant, 1 + 1 3 + 1 6 = 4 + 1 0 + 1 6 = 7 + 1 0 + 1 3 = 3 0 . If the the answer is 3 5 and the smallest n is 1 2 then the largest n would be 1 2 + 3 × 3 5 = 1 1 7 which is impossible. That is why we need only to check what is the maximum n is and it is 4 8 . Hafizh, if you physically count 1 2 , 1 5 , 1 8 , . . . , 4 8 , there are 1 3 and not 1 2 because we count 1 2 as 1 not 0 . For example, in the series { 1 , 2 , 3 } , there are 3 numbers not 3 − 1 = 2 numbers. You can also consider 1 2 = 3 ( 1 ) + 9 , 1 5 = 3 ( 2 ) + 9 , 1 8 = 3 ( 3 ) + 9 , . . . , 4 8 = 3 ( 1 3 ) + 9 , that is why there are 1 3 members.
and why the formula is (n(max)-n(min)/3)+1?
We note that the members of the set are of the form 3 i + 1 , where i = 0 , 1 , 2 , 3 , 4 , 5 , 6 . So the integer, which is the sum of three members, n = 3 ( i + j + k ) + 3 is a multiple of 3 .
We know that the minimum of such n , n m i n = 1 + 4 + 7 = 1 2 and the maximum, n m a x = 1 3 + 1 6 + 1 9 = 4 8 , then n ∈ { 1 2 , 1 5 , 1 8 , . . . , 4 8 } and there are 3 n m a x − n m i n + 1 = 3 4 8 − 1 2 + 1 = 1 3 of such n .