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If integer k

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by gmatblood » Mon Nov 07, 2011 9:50 pm
If integer k is equal to the sum of all even multiples of 15 between 295 and 615, what is the greatest prime factor of k?

5


7


11


13


17
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by shankar.ashwin » Mon Nov 07, 2011 9:55 pm
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
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by neelgandham » Tue Nov 08, 2011 1:44 am
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 !
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by saketk » Tue Nov 08, 2011 9:49 am
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.
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