15*20 = 300 and
15*41 = 615
Between 20 and 41 inclusive there are 11 even numbers,
15*(20+22+24+26+ ---- + 40)
15*2(10+11+12+13+ -- + 20)
15*2* (165).
Largest prime factor of 165 = 11.
C IMO
If integer k
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shankar.ashwin
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On the same lines as Shankar's Solution-
The Even factors of 15 between 295 and 615 are 300,330,360.......600. Sum of all the even factors is equal to 300+330+360+...600
= 300+330+360+...600
= 30*(10+11+12+13+14+15+16+17+18+19+20)
= 30*((20*21*0.5)-(9*10*0.5))
= 30*(210-45)
= 30*(11*15)
= 5*3*2*11*5*3
The greatest prime factor of k is 11, Ans C !
The Even factors of 15 between 295 and 615 are 300,330,360.......600. Sum of all the even factors is equal to 300+330+360+...600
= 300+330+360+...600
= 30*(10+11+12+13+14+15+16+17+18+19+20)
= 30*((20*21*0.5)-(9*10*0.5))
= 30*(210-45)
= 30*(11*15)
= 5*3*2*11*5*3
The greatest prime factor of k is 11, Ans C !
Anil Gandham
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saketk
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It's an Arithmetic Series
First term = 300
Last term = 600
Difference = 30
Use the sum of AP formula, you will get = 450*11 [ already factored, thanks to AP formula ]
I feel this is the quickest way to do this problem.
First term = 300
Last term = 600
Difference = 30
Use the sum of AP formula, you will get = 450*11 [ already factored, thanks to AP formula ]
I feel this is the quickest way to do this problem.












