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• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • ### Free GMAT Strategy Webinar November 17, 2018 November 17, 2018 07:00 AM PST 09:00 AM PST Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. # If integer k is equal to the sum of all even multiples of 15  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Senior Manager Joined: 31 Aug 2009 Posts: 386 Location: Sydney, Australia If integer k is equal to the sum of all even multiples of 15 [#permalink] ### Show Tags 11 Nov 2009, 04:27 3 15 00:00 Difficulty: 75% (hard) Question Stats: 55% (01:36) correct 45% (01:45) wrong based on 512 sessions ### HideShow timer Statistics 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 ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 50585 Re: Even multiples of 15 [#permalink] ### Show Tags 11 Nov 2009, 04:45 9 7 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 ANS = C Please explain how you arrive at your answer. Even multiple of 15 is of a form 15*2n=30n, hence it's is multiple of 30. First multiple of 30 in the range 295-615 is obviously 300 and the last 600. so we have: 300+330+360+...+600=30*(10+11+...+20) 10+11+..+20 sum of 11 consecutive integers =$$\frac{first \ term+last \ term}{2}*number \ of \ terms=\frac{10+20}{2}*11=15*11$$ $$300+330+360+...+600=30*(10+11+...+20)=30*15*11=2*3^2*5^2*11$$ Greatest prime 11. Answer: C (11) _________________ ##### General Discussion SVP Joined: 30 Apr 2008 Posts: 1830 Location: Oklahoma City Schools: Hard Knocks Re: Even multiples of 15 [#permalink] ### Show Tags 11 Nov 2009, 04:46 2 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.htm 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 ANS = C Please explain how you arrive at your answer. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Senior Manager Joined: 31 Aug 2009 Posts: 386 Location: Sydney, Australia Re: Even multiples of 15 [#permalink] ### Show Tags 11 Nov 2009, 04:55 awesome responses. +1 to both of you. Manager Joined: 28 Oct 2009 Posts: 79 Re: Number Properties [#permalink] ### Show Tags 17 May 2010, 15:46 Thanks, this was a difficult problem for me. Manager Joined: 25 Jul 2010 Posts: 105 Re: even multiples of 15 between 295 and 615 [#permalink] ### Show Tags 27 Sep 2010, 13:31 super. many thanks. Manager Joined: 16 Jun 2010 Posts: 150 Re: even multiples of 15 between 295 and 615 [#permalink] ### Show Tags 27 Sep 2010, 14:19 Bunnel --- Amazing solution. _________________ Please give me kudos, if you like the above post. Thanks. Manager Joined: 13 May 2011 Posts: 241 WE 1: IT 1 Yr WE 2: Supply Chain 5 Yrs Re: If integer k is equal to the sum of all even multiples of 15 [#permalink] ### Show Tags 12 Jan 2012, 04:44 2 300+330+.........+600 600=300+(n-1)*30 n=11 sum=k=11/2[2*300+10*30] =11*450= 11*5^2*3^2*2 the largest prime = 11 Manager Joined: 27 Oct 2011 Posts: 141 Location: United States Concentration: Finance, Strategy GPA: 3.7 WE: Account Management (Consumer Products) Re: If integer k is equal to the sum of all even multiples of 15 [#permalink] ### Show Tags 11 Feb 2012, 16:22 Those are great answers... i especially like the step of factoring out the 30. _________________ DETERMINED TO BREAK 700!!! Manager Joined: 27 Oct 2011 Posts: 141 Location: United States Concentration: Finance, Strategy GPA: 3.7 WE: Account Management (Consumer Products) [#permalink] ### Show Tags 11 Feb 2012, 16:25 even integers of 15 can only be multiples of 30. so find all the multiples of 30 in 300-600. 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. _________________ DETERMINED TO BREAK 700!!! Senior Manager Joined: 13 Oct 2016 Posts: 367 GPA: 3.98 Re: If integer k is equal to the sum of all even multiples of 15 [#permalink] ### Show Tags 26 Nov 2016, 07:36 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{(40-2)}{2} + 1 = 20$$ # of even numbers from 1 to 18 = $$\frac{(18-2)}{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$$. Current Student Status: DONE! Joined: 05 Sep 2016 Posts: 379 Re: If integer k is equal to the sum of all even multiples of 15 [#permalink] ### Show Tags 02 Dec 2016, 09:08 Great question to make you think!! multiples of 15 within the range --> 300 (15 x 20) ----- 615 (15 x 41) 41-20+1 = 22 multiples total 22/2 = 11 multiples that will be even median of 11 = 6 20+5 (we're including 20, so you'll have 6 values added together) = 25 15x25 = 375 (KEEP TO THE SIDE) sum/# = avg --> this nifty formula can be applied here! sum/11 = 375 --> sum = 11x375 Largest prime value is already in front of us --> 11! C. Math Expert Joined: 02 Aug 2009 Posts: 7030 Re: If integer k is equal to the sum of all even multiples of 15 between 2 [#permalink] ### Show Tags 29 Apr 2017, 19:20 1 dflorez wrote: If integer k is equal to the sum of all even multiples of 15 between 295 and 615, what is the greatest prime factor of k? A. 5 B. 7 C. 11 D. 13 E. 17 Even multiples of 15 means multiple of 30... 295 to 615 clearly shows 300-600.. So we are looking at 300+330+....+570+600... $$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. C _________________ 1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html GMAT online Tutor VP Joined: 05 Mar 2015 Posts: 1000 Re: If integer k is equal to the sum of all even multiples of 15 between 2 [#permalink] ### Show Tags 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 Ans C Current Student Joined: 12 Aug 2015 Posts: 2633 Schools: Boston U '20 (M) GRE 1: Q169 V154 Re: If integer k is equal to the sum of all even multiples of 15 between 2 [#permalink] ### Show Tags 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 SMASH THAT C. _________________ MBA Financing:- INDIAN PUBLIC BANKS vs PRODIGY FINANCE! Getting into HOLLYWOOD with an MBA! The MOST AFFORDABLE MBA programs! STONECOLD's BRUTAL Mock Tests for GMAT-Quant(700+) AVERAGE GRE Scores At The Top Business Schools! EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12856 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If integer k is equal to the sum of all even multiples of 15 [#permalink] ### Show Tags 26 Feb 2018, 11:33 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 _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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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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05 Aug 2018, 07:26
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 $$a_n=a+(n-1)d$$
$$40 =20+(n-1)2$$
$$n-1=\frac{20}{2}$$
$$n=11$$
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  [#permalink]

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06 Aug 2018, 08:17
Top Contributor
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 + 600
Since 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 + 20
Since 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

Cheers,
Brent
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Re: If integer k is equal to the sum of all even multiples of 15  [#permalink]

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21 Oct 2018, 05:12
Since any even number must be divisible by 2, any even multiple of 15 must be divisible by 2 and by 15, or in other words, must be divisible by 30. As a result, finding the sum of even multiples of 15 is equivalent to finding the sum of multiples of 30. By observation, the first multiple of 30 greater than 295 will be equal to 300 and the last multiple of 30 smaller than 615 will be equal to 600.

Thus, since there are no multiples of 30 between 295 and 299 and between 601 and 615, finding the sum of all multiples of 30 between 295 and 615, inclusive, is equivalent to finding the sum of all multiples of 30 between 300 and 600, inclusive. Therefore, we can rephrase the question: “What is the greatest prime factor of the sum of all multiples of 30 between 300 and 600, inclusive?”

The sum of a set = (the mean of the set) × (the number of terms in the set)

Since 300 is the 10th multiple of 30, and 600 is the 20th multiple of 30, we need to count all multiples of 30 between the 10th and the 20th multiples of 30, inclusive.
There are 11 terms in the set: 20th – 10th + 1 = 10 + 1 = 11
The mean of the set = (the first term + the last term) divided by 2: (300 + 600) / 2 = 450
k = the sum of this set = 450 × 11

Note, that since we need to find the greatest prime factor of k, we do not need to compute the actual value of k, but can simply break the product of 450 and 11 into its prime factors:
k = 450 × 11 = 2 × 3 × 3 × 5 × 5 × 11

Therefore, the largest prime factor of k is 11.

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Re: If integer k is equal to the sum of all even multiples of 15 &nbs [#permalink] 21 Oct 2018, 05:12
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