Mathematics
BernadetteVrbas
2016-04-27 09:26:27
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.
ANSWERS
4annanewton
2016-04-27 16:13:03

[latex]it a_{15} =126=3cdot42 \;\a_1=3n=? \;\ 3, 6, 9, 12, ... , 3n, ..., 3cdot42 [/latex] [latex]it 42-(n-1) =15 Rightarrow n-1=42-15=27 Rightarrow n=28 Rightarrow \;\ Rightarrow a_1=3cdot28= 84[/latex] Now, we have the sequence : [latex]it a_1= 3cdot28 = 84 otin M_9 \;\ a_2= 3cdot29 = 87 otin M_9 \;\ a_3= 3cdot30 = 90 in M_9 \;\ vdots[/latex] [latex]it a_{15} =3cdot42=126 in M_9 \;\ 90leq9cdot kleq126 |:9 Rightarrow 10leq kleq14 Rightarrow k in {10, 11, 12, 13, 14}[/latex] It implies 5 multiples of 9 At a glance :  {90,  99,  108,  117,  126}

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