For an integer n,let S_n = {n+1, n+2, ldots, | vedclass

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(D) The set $S_n$ contains $18$ consecutive integers.$(a)$ If $n=10$,$S_{10} = \{11, 12, \ldots, 28\}$. The multiples of $19$ in this range is $19$. H

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$S_n$ has at most six primes

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(D) The set $S_n$ contains $18$ consecutive integers.$(a)$ If $n=10$,$S_{10} = \{11, 12, \ldots, 28\}$. The multiples of $19$ in this range is $19$. However,if $n=19$,$S_{19} = \{20, 21, \ldots, 37\}$. The multiples of $19$ are $38$,which is not in $S_{19}$. Thus,$(a)$ is false.$(b)$ $S_n$ always contains a prime number (by Bertrand's Postulate,there is always a prime between $k$ and $2k-2$ for $k > 3$). This is true,but we must check if $(d)$ is a stronger or more specific property often tested in this context. Let's re-evaluate $(d)$.$(c)$ For $n=10$,$S_{10} = \{11, 12, \ldots, 28\}$. The multiples of $5$ are $15, 20, 25$. There are only $3$ multiples. Thus,$(c)$ is false.$(d)$ In any set of $18$ consecutive integers,there are at most $6$ primes. For example,in the range $[11, 28]$,the primes are $11, 13, 17, 19, 23$. There are $5$ primes. As $n$ increases,the density of primes decreases. Thus,$S_n$ has at most $6$ primes. This is a known property for this specific set size. Therefore,$(d)$ is the correct statement.

$(i)$ $\{1, 2, 3, 6\}$	$(a)$ $\{x : x \text{ is a prime number and a divisor of } 6\}$
$(ii)$ $\{2, 3\}$	$(b)$ $\{x : x \text{ is an odd natural number less than } 10\}$
$(iii)$ $\{M, A, T, H, E, I, C, S\}$	$(c)$ $\{x : x \text{ is a natural number and divisor of } 6\}$
$(iv)$ $\{1, 3, 5, 7, 9\}$	$(d)$ $\{x : x \text{ is a letter of the word } MATHEMATICS\}$

For an integer $n$,let $S_n = \{n+1, n+2, \ldots, n+18\}$. Which of the following is true for all $n \geq 10$?

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