Q36 / 60 · KVPY 2012 (SA) · Mathematics

Let $S=\{1,2,3, \ldots . ., 40\}$ and let $A$ be a subset of $S$ such that no two elements in $A$ have their sum divisible by 5 . What is the maximum number of elements possible in $\mathrm{A}$ ?

  • A
    10
  • B
    13
  • Correct answer
    C
    17
  • D
    20

Solution

$\mathrm{A}=\{1,2,6,7,11,12,16,17,21,22,26,27,31,32,36,37\} \&$ One of the element which is multiple of 5 $\mathrm{B}=\{3,4,8,9,13,14,18,19,23,24,28,29,33,34,38,39\} \&$ One of the element which is multiple of 5

Hence, the correct answer is option: 'C'.

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