Question

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. (2) the greatest term of s is 126.

Answer 1

$\it a_{15} =126=3\cdot42 \\\;\\a_1=3n=? \\\;\\ 3, 6, 9, 12, ... , 3n, ..., 3\cdot42$

$\it 42-(n-1) =15 \Rightarrow n-1=42-15=27 \Rightarrow n=28 \Rightarrow \\\;\\ \Rightarrow a_1=3\cdot28= 84$

Now, we have the sequence :

$\it a_1= 3\cdot28 = 84 \notin M_9 \\\;\\ a_2= 3\cdot29 = 87 \notin M_9 \\\;\\ a_3= 3\cdot30 = 90 \in M_9 \\\;\\ \vdots$

$\it a_{15} =3\cdot42=126 \in M_9 \\\;\\ 90\leq9\cdot k\leq126 |:9 \Rightarrow 10\leq k\leq14 \Rightarrow k \in \{10, 11, 12, 13, 14\}$

It implies 5 multiples of 9

At a glance :  {90,  99,  108,  117,  126}