If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?

(1) There are 15 terms in S.

In every set of 3 consecutive multiples of 3, there will be one multiple of 9:
{-9, -6, -3}
{-6, -3, 0}
{-3, 0, 3}
{0, 3, 6}
{3, 6, 9}
...

15 consecutive multiples of 3 could be break down into 5 different such sets, each of which will contain 1 multiple of 9. Therefore, there are 5 multiples of 9 in the set. Sufficient.

(2) The greatest term of S is 126. Clearly insufficient.

Answer: A.
_________________

Asked: If S is a sequence of consecutive multiples of 3, how many multiples of 9 are there in S?

Let S = 3k, 3(k+1),...

(1) There are 15 terms in S.
S = 3k, 3(k+1), .... 3(k+14)
If k is a multiple of 3; multiple of 9 = 3k, 3(k+3), 3(k+6), 3(k+9), 3(k+12); Total 5 multiple of 9.
If k+1 is a multiple of 3; multiple of 9 = 3(k+1), 3(k+4), 3(k+7), 3(k+10), 3(k+13); Total 5 multiple of 9.
If k+2 is a multiple of 3: multiple of 9 = 3(k+2), 3(k+5), 3(k+8), 3(k+11), 3(k+14); Total 5 multiple of 9.
There is no other possibility.
There are 5 multiples of 9.
SUFFICIENT

(2) The greatest term of S is 126.
Since the number of terms are not known
NOT SUFFICIENT

IMO A
_________________