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If integer k is equal to the sum of all even multiples of 15
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11 Nov 2009, 05:27
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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
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Re: Even multiples of 15
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11 Nov 2009, 05:45




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Re: Even multiples of 15
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11 Nov 2009, 05:46
C. You can find the sum by taking the average of the highest and lowest numbers that satisfy your criteria multiplied by how many numbers there are. Find out how many numbers there are by subtracting the highest number that satisfies your condition minus the lowest number that satisfies your condiction, divide by the interval (here that's 30), then add 1. 600  300 = 300; 300 / 30 = 10, 10 +1 = 11 Now the average of 600 and 300 is 450. so 450 * 11 = 4950. Notice also that 11 is prime and the question calls for the largest prime number. Because we already used 11, we know C is possible, and 13 and 17 are not factors. 4950 / 13 = 380 and 10/13  not a factor 4950 / 17 = 291 and 3/17 = not a factor. So C is the answer. For methods on finding divisibility by the primes up to 50, check out this link http://home.egge.net/~savory/maths1.htmyangsta8 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 Please explain how you arrive at your answer.
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Re: Even multiples of 15
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11 Nov 2009, 05:55
awesome responses. +1 to both of you.



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17 May 2010, 16:46
Thanks, this was a difficult problem for me.



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Re: even multiples of 15 between 295 and 615
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27 Sep 2010, 14:31
super. many thanks.



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Re: even multiples of 15 between 295 and 615
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27 Sep 2010, 15:19
Bunnel  Amazing solution.
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Re: If integer k is equal to the sum of all even multiples of 15
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12 Jan 2012, 05:44
300+330+.........+600 600=300+(n1)*30 n=11
sum=k=11/2[2*300+10*30] =11*450= 11*5^2*3^2*2 the largest prime = 11



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Re: If integer k is equal to the sum of all even multiples of 15
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11 Feb 2012, 17:22
Those are great answers... i especially like the step of factoring out the 30.
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even integers of 15 can only be multiples of 30. so find all the multiples of 30 in 300600. then you can factor out and find that the total is 4950 and there is a factor of 11 in there which makes it largest prime. factor.
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Re: If integer k is equal to the sum of all even multiples of 15
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26 Nov 2016, 08:36
yangsta8 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 Slightly different approach. We have \(15*20 + 15*22 + 15*24 + ….. + 15*40 = 15*(20 + 22 + 24 + … + 40)\) = = 15* [(sum of even numbers from 1 to 40) – (sum of even numbers from 1 to 18)]# of even numbers from 1 to 40 = \(\frac{(402)}{2} + 1 = 20\) # of even numbers from 1 to 18 = \(\frac{(182)}{2} + 1 = 9\) Sum of consecutive even numbers = \(n(n+1)\) \(= 15*(20*21  9*10) = 15*(420 – 90) = 15*330=15*30*11\) Greatest prime factor is \(11\).



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Re: If integer k is equal to the sum of all even multiples of 15
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02 Dec 2016, 10:08
Great question to make you think!! multiples of 15 within the range > 300 (15 x 20)  615 (15 x 41) 4120+1 = 22 multiples total 22/2 = 11 multiples that will be even median of 11 = 6 20+5 (we're including 20, so you'll have 6 values added together) = 25 15x25 = 375 (KEEP TO THE SIDE) sum/# = avg > this nifty formula can be applied here! sum/11 = 375 > sum = 11x375 Largest prime value is already in front of us > 11! C.



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Re: If integer k is equal to the sum of all even multiples of 15 between 2
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29 Apr 2017, 20:20
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 300600.. 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. C
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Re: If integer k is equal to the sum of all even multiples of 15 between 2
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29 Apr 2017, 20:40
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 Ans C



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Re: If integer k is equal to the sum of all even multiples of 15 between 2
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29 Apr 2017, 21:07
Even multiple of 15 basically means multiples of 30 Series => 300,330....600 Mean => 900/2=450 Number of terms => 600300/30 +1 => 10+1 => 11 Sum => 11*450 => 11*5^2*3^2*2 Hence k=> 2*3^2*5^2*11 Greatest Prime => 11 SMASH THAT C.
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Re: If integer k is equal to the sum of all even multiples of 15
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26 Feb 2018, 12:33
Hi All, The "math" behind this question can be done in a variety of ways, but you will eventually have to use what's called "prime factorization" to answer this question. First, we need to figure out what we're summing: the even multiples of 15 between 295 and 615. That will be all the even multiples of 15 from 300 to 600, inclusive. 300 + 330 + 360…..+600 Since these numbers are all multiples of 30, we have…. 30(10 + 11 + 12…..+20) Using "bunching" rules, this gives us…. 30(165) Prime factoring this gives us… (3x2x5)(11x3x5) So the greatest prime factor is 11 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If integer k is equal to the sum of all even multiples of 15 between 2
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05 Aug 2018, 08:26
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 OA: C Even multiples of \(15\) between \(295\) and \(615\) are \(300(20*15),330(22*15),360(24*15),................................,600(40*15)\) \(K =\) sum of even multiples of \(15\) between \(295\) and \(615\) \(= 300+ 330+360+...........................+600\) \(K=15( 20+22+24+26+....................+40)\) \(20+22+24+26+....................+40\) is sum of an A.P series with \(a=20\) and \(d=2\) to find the number of terms in A.P, we will use formula \(a_n=a+(n1)d\) \(40 =20+(n1)2\) \(n1=\frac{20}{2}\) \(n=11\) Sum of A.P series \(= \frac{n}{2}\)[First Term + Last Term]\(=\frac{11}{2}[20+40]=11*30\) \(K =15*11*30=2^1*3^2*5^2*11^1\) Greatest prime factor of \(K =11\)
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Re: If integer k is equal to the sum of all even multiples of 15
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06 Aug 2018, 09:17
yangsta8 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 Multiples of 15: 15, 30, 45, 60, 75, 90, 105, etc EVEN multiples of 15: 30, 60, 90, 120, .... So k = 300 + 330 + 360 + ... + 570 + 600Since each number is a multiple of 30, let's rewrite each value as the product of 30 and some integer: 300 = 30( 10) 330 = 30( 11) 360 = 30( 12) 390 = 30( 13) . . . 570 = 30( 19) 600 = 30( 20) So k = 30( 10 + 11 + 12 + ... + 19 + 20)  Let's examine this sum: 10 + 11 + 12 + ... + 19 + 20Since 20  10 + 1 = 11, we know there are 11 numbers to add together. Since these red numbers are equally spaced (consecutive integers), their sum = (# of values)(average of first and last values) = [ 11][(10+20)/2] = [ 11][15] = (11)(15) So, k = 30( 10 + 11 + 12 + ... + 19 + 20) = 30 (11)(15)= (2)(3)(5) (11)(3)(5)We can see that 11 is the greatest prime factor of k Answer:C Cheers, Brent
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