Consider the sets
$T_{n} = n, n + 3, n + 5, n + 7, n + 9$
where n = 1, 2, 3,..., 99. How many of these sets contain 5 or any integral multiple thereof (i.e. any one of the numbers 5, 10, 15, 20, 25,...)?

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none of these

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Solution

The correct option is B 79
$\begin{matrix} T_{n} = n & n + 3 & n + 5 & n + 7 & n + 9 T_{1} = 1 & 4 & 6 & 8 & 10 T_{2} = 2 & 5 & 7 & 9 & 11 T_{3} = 3 & 6 & 8 & 10 & 12 T_{4} = 4 & 7 & 9 & 11 & 13 T_{5} = 5 & 8 & 10 & 12 & 14 T_{6} = 6 & 9 & 11 & 13 & 15 T_{7} = 7 & 10 & 12 & 14 & 16 T_{8} = 8 & 11 & 13 & 15 & 17 T_{9} = 9 & 12 & 14 & 16 & 18 \end{matrix}$
In general when n = 4, 9, 14, 19,..., 99 does not contain 5 or its multiple
i.e. n = 5 k - 1 where k = 1, 2, 3..., 20
Hence out of 99 sets, 20 sets do not contain 5 or its multiples.
Thus the required number of sets = 99 - 20 = 79

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