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If integer k is equal to the sum of all even multiples of 15 between 2

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If integer k is equal to the sum of all even multiples of 15 between 2 [#permalink]

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New post 29 Apr 2017, 15:07
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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?

A. 5
B. 7
C. 11
D. 13
E. 17
[Reveal] Spoiler: OA
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Re: If integer k is equal to the sum of all even multiples of 15 between 2 [#permalink]

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New post 29 Apr 2017, 19:20
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dflorez wrote:
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?

A. 5
B. 7
C. 11
D. 13
E. 17


Even multiples of 15 means multiple of 30...
295 to 615 clearly shows 300-600..
So we are looking at 300+330+....+570+600...
\(300+330+360+....+570+600=30(10+11+12+....+19+20)\)
Now 10+11+12+.....+20= average of lowest and greatest * number of items=\(\frac{10+20}{2}*11=15*11\)..
So sum is 30*15*11....
Clearly 11 is the biggest prime factor.

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Re: If integer k is equal to the sum of all even multiples of 15 between 2 [#permalink]

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New post 29 Apr 2017, 19:40
dflorez wrote:
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?

A. 5
B. 7
C. 11
D. 13
E. 17



sum = mean *no. of terms
since all are in sequence , mean = 300+600/2= 450
no of terms = 15*2 (10,11,12....20) , total 11 nos.
thus Sum = 450*11 = 9*5*5*2*11
clearly largest prime is 11

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Re: If integer k is equal to the sum of all even multiples of 15 between 2 [#permalink]

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New post 29 Apr 2017, 20:07
Even multiple of 15 basically means multiples of 30

Series => 300,330....600
Mean => 900/2=450
Number of terms => 600-300/30 +1 => 10+1 => 11

Sum => 11*450 => 11*5^2*3^2*2
Hence k=> 2*3^2*5^2*11

Greatest Prime => 11

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Re: If integer k is equal to the sum of all even multiples of 15 between 2   [#permalink] 29 Apr 2017, 20:07
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