Answer:
The greatest prime factor of k is 11.
Step-by-step explanation:
∵ LCM( 2, 15) = 30,
Thus, the even multiple of 15 must be multiple of 30,
That is, the even multiple of 15 between 295 and 615 are
300, 330, 360,..........600
Which is an AP,
Having first term, a = 300,
Common difference, d = 30,
If n be the number terms,
Last term = [tex]a+(n-1)d[/tex]
[tex]=300+(n-1)30[/tex]
[tex]=270 + 30n[/tex]
[tex]\implies 270+30n = 600\implies 30n = 330\implies n = 11[/tex]
Hence, the sum of the all even multiple of 15 from 295 to 615,
[tex]S_{11}=\frac{11}{2}(2(300)+(11-1)30)=\frac{11}{2}(600+300)=\frac{11}{2}(900)=11\times 450[/tex]
According to the question,
[tex]S_{11}=k[/tex]
⇒ k = 11 × 450 = 11 × 5 × 5 × 3 × 3 × 2
Hence, the greatest prime factor of k is 11.
