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
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Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com